Exponents and Polynomials

Transcription

1 C H A P T E R 5 Eponents nd Polynomils ne sttistic tht cn be used to mesure the generl helth of ntion or group within ntion is life epectncy. This dt is considered more ccurte thn mny other sttistics becuse it is esy to determine the precise number of yers in person s lifetime. According to the Ntionl Center for Helth Sttistics, n Americn born in 996 hs life epectncy of 76. yers. However, n Americn mle born in 996 hs life epectncy of only 7.0 yers, wheres femle cn epect 79.0 yers. A mle who mkes it to 65 cn epect to live 5.7 more yers, wheres femle who mkes it to 65 cn epect 8.9 more yers. In the net few yers, thnks in prt to dvnces in helth cre nd science, longevity is epected to increse significntly worldwide. In fct, the World Helth Orgniztion predicts tht by 05 no country will hve life epectncy of less thn 50 yers. In this chpter we study lgebric epressions involving eponents. In Eercises 95 nd 96 of Section 5. you will see how formuls involving eponents cn be used to find the life epectncies of men nd women. O 80 U.S. fem U.S les. m les Yer of birth Life epectncy (yers) y

2 dugoi_c05.qd_a 0/5/00 58 (5-) 0: AM Chpter 5 Pge 58 Eponents nd Polynomils INTEGRAL EXPONENTS AND SCIENTIFIC NOTATION 5. In Chpter we defined positive integrl eponents nd lerned to evlute epressions involving eponents. In this section we will etend the definition of eponents to include ll integers nd to lern some rules for working with integrl eponents. In Chpter 7 we will see tht ny rtionl number cn be used s n eponent. In this section Positive nd Negtive Eponents Product Rule Zero Eponent Chnging the Sign of n Eponent Quotient Rule Scientific Nottion Positive nd Negtive Eponents Positive integrl eponents provide convenient wy to write repeted multipliction or very lrge numbers. For emple,, y y y y y,,000,000, nd We refer to s cubed, rised to the third power, or power of. Positive Integrl Eponents If is nonzero rel number nd n is positive integer, then n.... n fctors of n In the eponentil epression, the bse is, nd the eponent is n. We use to represent the reciprocl of. Becuse 8, we hve 8. In generl, n is defined s the reciprocl of n. Negtive Integrl Eponents If is nonzero rel number nd n is positive integer, then n n. (If n is positive, n is negtive.) To evlute, you cn first cube to get 8 nd then find the reciprocl to get, or you cn first find the reciprocl of which is nd then cube to get helpful. 8 hint A negtive eponent does not cuse n epression to hve negtive vlue. The negtive eponent cuses the reciprocl: 8 So. The power nd the reciprocl cn be found in either order. If the eponent is, we simply find the reciprocl. For emple,,, 8 ( ), 8, nd. Becuse nd re reciprocls of ech other, we hve ( ). 6 nd. These emples illustrte the following rules.

3 dugoi_c05.qd_a 0/0/00 5:8 PM Pge Integrl Eponents nd Scientific Nottion (5-) 59 s Rules for Negtive Eponents If is nonzero rel number nd n is positive integer, then n n, Negtive eponents Evlute ech epression. b) ( ) ) clcultor, n n. nd ) 9 b) ( ) ( ) 9 c) 9 d) 6 7 e) close-up You cn evlute epressions with negtive eponents using grphing clcultor. Use the frction feture to get frctionl nswers. d) c) e) 5 Definition of negtive eponent Definition of negtive eponent Evlute, then tke the opposite. The reciprocl of is. 6 The cube of is. 7 The reciprocl of 5 is 5. In Chpter we greed to evlute by squring first nd then tking the opposite. So 9, wheres ( ) 9. The sme greement lso holds for negtive eponents. Tht is why the nswer to Emple (c) is negtive. CAUTION clcultor Product Rule We cn simplify n epression such s 5 using the definition of eponents. close-up three s five s A grphing clcultor cnnot prove tht the product rule is correct, but it cn provide numericl support for the product rule. ( )( ) 8 5 eight s Notice tht the eponent 8 is the sum of the eponents nd 5. This emple illustrtes the product rule for eponents. Product Rule for Eponents If m nd n re integers nd 0, then m n m n. Using the product rule Simplify ech epression. Write nswers with positive eponents nd ssume ll vribles represent nonzero rel numbers. b) 5 c) y ( 5y ) ) 6

4 60 (5-) Chpter 5 helpful Eponents nd Polynomils ) Product rule b) Product rule: 0 Definition of negtive eponent c) y ( 5y ) ( )( 5)y y 0y 7 Product rule: ( ) Definition of negtive eponent y hint The definitions of the different types of eponents re relly clever mthemticl invention. The fct tht we hve rules for performing rithmetic with those eponents mkes the nottion of eponents even more mzing. The product rule cnnot be pplied to becuse the bses re not identicl. Even when the bses re identicl, we do not multiply the bses. For emple, 5 9. Using the rule correctly, we get 5 9. CAUTION Zero Eponent We hve used positive nd negtive integrl eponents, but we hve not yet seen the integer 0 used s n eponent. Note tht the product rule ws stted to hold for ny integers m nd n. If we use the product rule on, we get 0. Becuse 8 nd 8, we must hve. So for consistency we define 0 nd the zero power of ny nonzero number to be. Zero Eponent If is ny nonzero rel number, then 0. Using zero s n eponent Simplify ech epression. Write nswers with positive eponents nd ssume ll vribles represent nonzero rel numbers. 0 b) c) 5b 6 5b ) 0 ) To evlute 0, we find 0 nd then tke the opposite. So 0. 0 b) Definition of zero eponent c) 5b 6 5b 65 5 b 6 b 60b Product rule helpful hint Defining 0 to be gives consistent pttern to eponents: 9 6 b 0 9 Definitions of negtive nd zero eponent Chnging the Sign of n Eponent Becuse n nd n re reciprocls of ech other, we know tht n n If the eponent is incresed by (with bse ) the vlue of the epression is multiplied by. nd n. n

5 5. Integrl Eponents nd Scientific Nottion (5-5) 6 So negtive eponent in the numertor or denomintor cn be chnged to positive by relocting the eponentil epression. In the net emple we use these fcts to remove negtive eponents from eponentil epressions. Simplifying epressions with negtive eponents Write ech epression without negtive eponents nd simplify. All vribles represent nonzero rel numbers. 5 ) b) y z 5 ) Rewrite division s multipliction. 5 5 Chnge the signs of the negtive eponents. 0 5 Product rule: 5 Note tht in 5 the negtive eponent pplies only to. b) Rewrite s multipliction. y z y z y Definition of negtive eponent z y Simplify. z In Emple we showed more steps thn re necessry. For instnce, in prt (b) we could simply write y. y z z Eponentil epressions (tht re fctors) cn be moved from numertor to denomintor (or vice vers) s long s we chnge the sign of the eponent. clcultor If n eponentil epression is not fctor, you cnnot move CAUTION it from numertor to denomintor (or vice vers). For emple,. close-up Becuse nd, we get A grphing clcultor cnnot prove tht the quotient rule is correct, but it cn provide numericl support for the quotient rule. not. Quotient Rule We cn use rithmetic to simplify the quotient of two eponentil epressions. For emple, 5.

6 6 (5-6) Chpter 5 Eponents nd Polynomils There re five s in the numertor nd three s in the denomintor. After dividing, two s remin. The eponent in cn be obtined by subtrcting the eponents nd 5. This emple illustrtes the quotient rule for eponents. Quotient Rule for Eponents If m nd n re ny integers nd 0, then m n m n. study 5 Using the quotient rule Simplify ech epression. Write nswers with positive eponents only. All vribles represent nonzero rel numbers. 9 y m5 b) c) ) y m 9 ) 9 5 tip The keys to college success re motivtion nd time mngement. Students who tell you tht they re mking gret grdes without studying re probbly not telling the truth. Success in college tkes effort. m5 m b) m5 ( ) m8 y y y c) y ( ) y The net emple further illustrtes the rules of eponents. Remember tht the bses must be identicl for the quotient rule or the product rule. 6 Using the product nd quotient rules Use the rules of eponents to simplify ech epression. Write nswers with positive eponents only. All vribles represent nonzero rel numbers. 7 w(w ) y5 b) c) ) 7 w y 7 ) 0 Quotient rule: 7 ( 7) 0 7 Definition of zero eponent w(w ) w b) Product rule: w w w w w w Quotient rule: ( ) Definition of negtive eponent w y5 y5 y c) y y y

7 5. Integrl Eponents nd Scientific Nottion (5-7) 6 Scientific Nottion Mny of the numbers tht re encountered in science re either very lrge or very smll. For emple, the distnce from the erth to the sun is 9,000,000 miles, nd hydrogen tom weighs grm. Scientific nottion provides convenient wy of writing very lrge nd very smll numbers. In scientific nottion the distnce from the erth to the sun is miles nd hydrogen tom weighs.7 0 grm. In scientific nottion the times symbol,, is used to indicte multipliction. Converting number from scientific nottion to stndrd nottion is simply mtter of multipliction. 7 Scientific nottion to stndrd nottion Write ech number using stndrd nottion. b) ) clcultor ) Multiplying number by 05 moves the deciml point five plces to the right: ,000 close-up b) Multiplying number by 0 or moves the deciml point four plces to the left: In norml mode, disply number in scientific nottion nd press ENTER to convert to stndrd nottion. You cn use power of 0 or the EE key to get the E for the built-in scientific nottion. The procedure for converting number from scientific nottion to stndrd nottion is summrized s follows. Strtegy for Converting from Scientific Nottion to Stndrd Nottion. Determine the number of plces to move the deciml point by emining the eponent on the 0.. Move to the right for positive eponent nd to the left for negtive eponent. A positive number in scientific nottion is written s product of number between nd 0, nd power of 0. Numbers in scientific nottion re written with only one digit to the left of the deciml point. A number lrger thn 0 is written with positive power of 0, nd positive number smller thn is written with negtive power of 0. Numbers between nd 0 re usully not written in scientific nottion. To convert to scientific nottion, we reverse the strtegy for converting from scientific nottion. Strtegy for Converting from Stndrd Nottion to Scientific Nottion. Count the number of plces (n) tht the deciml point must be moved so tht it will follow the first nonzero digit of the number.. If the originl number ws lrger thn 0, use 0n.. If the originl number ws smller thn, use 0 n.

8 6 (5-8) Chpter 5 Eponents nd Polynomils 8 Stndrd nottion to scientific nottion Convert ech number to scientific nottion. ) 9,000,000 b) clcultor close-up ) In 9,000,000 the deciml point must be moved eight plces to the left to get it to follow 9, the first nonzero digit. To convert stndrd nottion to scientific nottion, disply the number with the clcultor in scientific mode (Sci) nd then press ENTER. In scientific mode ll results re given in scientific nottion. 9,000, Use 8 becuse 9,000, b) The deciml point in must be moved si plces to the right to get the to the left of the deciml point Use 6 becuse We cn perform computtions with numbers in scientific nottion by using the rules of eponents on the powers of 0. 9 Using scientific nottion in computtions Evlute (0,000)( ) by first converting ech number to scientific nottion. (0,000)( ) ( 0)(.5 0 5) Commuttive nd ssocitive properties Write 0.5 in scientific nottion. clcultor close-up If you use powers of 0 to perform the computtion in Emple 9, you will need prentheses s shown. If you use the built-in scientific nottion you don t need prentheses.

9 5. helpful 0 hint (5-9) Integrl Eponents nd Scientific Nottion 65 Counting hydrogen toms If the weight of hydrogen is.7 0 grm per tom, then how mny hydrogen toms re there in one kilogrm of hydrogen? There re 000 or 0 grms in one kilogrm. So to find the number of hydrogen toms in one kilogrm of hydrogen, we divide 0 by.7 0 : You cn divide 0 by.7 0 without clcultor by dividing by.7 to get 0.59 nd 0 by 0 to get 07.Then convert to or g/kg tom per kilogrm (tom/kg).7 0 g/tom tom m To divide by grms per tom, we invert nd multiply: g to. Keeping kg kg g trck of the units s we did here helps us to be sure tht we performed the correct opertion. So there re pproimtely hydrogen toms in one kilogrm of WARM-UPS True or flse? Eplin your nswer Flse True. True Flse Flse 7. True True True True Flse 5 EXERCISES Reding nd Writing After reding this section, write out the nswers to these questions. Use complete sentences.. Wht is n eponentil epression? 6. How do you convert number from stndrd nottion to scientific nottion? To convert from stndrd nottion to scientific nottion, count the number of deciml plces, n, required to move the. Wht is the mening of negtive eponent? n nd 0.. Wht is the product rule? the right. For ll eercises in this section, ssume tht the vribles represent nonzero rel numbers nd use only positive eponents in your nswers. Evlute ech epression. See Emple Wht is the quotient rule? The quotient rule sys tht 5. How do you convert number from scientific nottion to stndrd nottion? To convert number in scientific nottion to stndrd nottion, move the deciml point n plces to the left if the eponent on 0 is n or move the deciml point n plces to the right if the eponent on 0 is n, ssuming n is positive integer

10 dugoi_c05.qd_a 0/5/00 66 (5-0) 0:6 AM Chpter 5 Pge 66 Eponents nd Polynomils ( 6) , ( ) , Simplify. See Emples nd y y y ( y) 8. ( 8 8)0 0. ( y)0. w (w7 w ). 5yz(y z ) Write ech epression without negtive eponents nd simplify. See Emple y y () 5 y y 6 y y yy 7 Simplify ech epression. See Emples 5 nd w w 5 w w w w 5ww 7 6 y r t y 0r5t t Use the rules of eponents to simplify ech epression ( ) ( ) ( ) 59. ( ) 60. ( ) y 5 5( ) 6( b ) b b 6 ( y)( y ) 67. 9y 5 ( 5y)( y6) y ( ) ( ) For ech eqution, find the integer tht cn be used s the eponent to mke the eqution correct. 7. 7? 7. 8? 75.?? 77. 6? ? 5? ? 0,000 Write ech number in stndrd nottion. See Emple ,000, ,000,000 96, Write ech number in scientific nottion. See Emple , ,98, ,95, Evlute ech epression using scientific nottion without clcultor. See Emple 9. (5,000,000)(0.000) (6000)(0.0000) (0,000)(0.00) 7.5

11 5. Integrl Eponents nd Scientific Nottion ( 05)(6 0 9) (.8 0 )(5 0 8) 0. (. 0 6)( 0) 5 Wste per person per dy (pounds) 0. (. 0 )(8.7 0 ) (.7 0 ) ( ) 06. (6.7 05) ( ) ( )(6. 076) 08. (.5 0 )(.7 0 ) Solid wste per person. In 960 the people in the United Sttes generted tons of municipl solid wste (Environmentl Protection Agency, How mny pounds per person per dy were generted in 960?.6 lb/person/dy Evlute ech epression using clcultor. Write nswers in scientific nottion. Round the deciml prt to three deciml plces. See Emple (. 09)( ) 6 (5-) Number of yers fter FIGURE FOR EXERCISES 5 AND 6 (5.6 0)(. 0 6) 09. (6. 0 ) (.5 0 6)(000) An incresing problem. According to the EPA, in 998 the people in the United Sttes generted. 0 pounds of solid municipl wste. ) How mny pounds per person per dy were generted in 998?. lb/person/dy in 998 b) Use the grph to predict the number of pounds per person per dy tht will be generted in the yer lb/person/dy in Solve ech problem.. Distnce to the sun. The distnce from the erth to the sun is 9 million miles. Epress this distnce in feet using scientific nottion ( mile 5,80 feet)..90 feet GET TING MORE INVOLVED 7. Eplortion. ) Using pirs of integers, find vlues for m nd n for which m n 6m n. b) For which vlues of m nd n is it true tht m n 6m n? 9 million miles Erth Sun 8. Coopertive lerning. Work in group to find the units digit of 99 nd eplin how you found it. FIGURE FOR EXERCISE 9. Discussion. Wht is the difference between n nd ( )n, where n is n integer? For which vlues of nd n do they hve the sme vlue, nd for which vlues of nd n do they hve different vlues?. Trveling time. The speed of light is feet per second. How long does it tke light to get from the sun to the erth? (See Eercise.) 8. minutes 0. Eplortion. If b, then wht cn you conclude bout b? Use scientific nottion on your clcultor to find Eplin why your clcultor displys the nswer tht it gets.. Spce trvel. How long does it tke spcecrft trveling. 05 kilometers per second to trvel.6 0 kilometers?.8 0. Dimeter of dot. If the circumference of very smll circle is meter, then wht is the dimeter of the circle?

12 68 (5-) Chpter 5 Eponents nd Polynomils 5. In Section 5. you lerned some of the bsic rules for working with eponents. All of the rules of eponents re designed to mke it esier to work with eponentil epressions. In this section we will etend our list of rules to include three new ones. In this section Rising n Eponentil Epression to Power Rising Product to Power Rising Quotient to Power Vrible Eponents Summry of the Rules Applictions THE POWER RULES Rising n Eponentil Epression to Power An epression such s ( ) consists of the eponentil epression rised to the power. We cn use known rules to simplify this epression. ( ) 6 Eponent indictes two fctors of. Product rule: 6 Note tht the eponent 6 is the product of the eponents nd. This emple illustrtes the power of power rule. Power of Power Rule If m nd n re ny integers nd 0, then (m)n mn. Using the power of power rule Use the rules of eponents to simplify ech epression. Write the nswer with positive eponents only. Assume ll vribles represent nonzero rel numbers. b) () 6 ) ()5 ( ) c) (y ) y 5 d) ( ) clcultor close-up ) ()5 5 Power of power rule 6 b) ( ) Power of power rule Definition of negtive eponent c) (y ) y 5 y6y 5 Power of power rule y Product rule ( ) d) Power of power rule ( ) 9 7 Quotient rule A grphing clcultor cnnot prove tht the power of power rule is correct, but it cn provide numericl support for it.

13 dugoi_c05.qd_a 0/5/00 0:7 AM Pge The Power Rules (5-) 69 Rising Product to Power clcultor Consider how we would simplify product rised to positive power nd product rised to negtive power using known rules. close-up fctors of () 8 (y) y (y) (y)(y)(y) y You cn use grphing clcultor to illustrte the power of product rule. In ech of these cses the originl eponent is pplied to ech fctor of the product. These emples illustrte the power of product rule. Power of Product Rule If nd b re nonzero rel numbers nd n is ny integer, then (b)n n bn. Using the power of product rule Simplify. Assume the vribles represent nonzero rel numbers. Write the nswers with positive eponents only. b) () c) (y) ) () ) () () Power of product rule 8 b) ( ) ()( ) Power of product rule 8 6 Power of power rule c) (y) ()()(y) y y Rising Quotient to Power clcultor Now consider n emple of pplying known rules to power of quotient: close-up We get similr result with negtive power: You cn use grphing clcultor to illustrte the power of quotient rule In ech of these cses the originl eponent pplies to both the numertor nd denomintor. These emples illustrte the power of quotient rule. Power of Quotient Rule If nd b re nonzero rel numbers nd n is ny integer, then b n n n. b

14 70 (5-) Chpter 5 Eponents nd Polynomils Using the power of quotient rule Use the rules of eponents to simplify ech epression. Write your nswers with positive eponents only. Assume the vribles re nonzero rel numbers. b) c) d) ) y helpful ) Power of quotient rule 8 ( ) 9 b) Becuse ( ) 9 nd (y) y6 y y y 7y ( ) 6 66 c) 8 d) 6 ( ) 9 hint The eponent rules in this section pply to epressions tht involve only multipliction nd division. This is not too surprising since eponents, multipliction, nd division re closely relted. Recll tht nd b b. A frction to negtive power cn be simplified by using the power of quotient rule s in Emple. Another method is to find the reciprocl of the frction first, then use the power of quotient rule s shown in the net emple. Negtive powers of frctions Simplify. Assume the vribles with positive eponents only. b) ) re nonzero rel numbers nd write the nswers 5 ) The reciprocl of is. Power of quotient rule b) 5 ( ) c) y y c) y 9 6 y Vrible Eponents So fr, we hve used the rules of eponents only on epressions with integrl eponents. However, we cn use the rules to simplify epressions hving vrible eponents tht represent integers. 5 Epressions with vribles s eponents Simplify. Assume the vribles represent integers. b) (5 ) ) y 5y n c) m 5n

15 5. ) y 5y 9y b) (5) 56 (n)5n n 5n c) m (m)5n 5n 5mn clcultor close-up Did we forget to include the rule ( b)n n bn? You cn esily check with clcultor tht this rule is not correct. The Power Rules (5-5) 7 Product rule: y 5y 9y Power of power rule: 6 Power of quotient rule Power of power rule Summry of the Rules The definitions nd rules tht were introduced in the lst two sections re summrized in the following bo. Rules for Integrl Eponents For these rules m nd n re integers nd nd b re nonzero rel numbers.. n n Definition of negtive eponent n. n,, nd n Negtive eponent rules n. 0 Definition of zero eponent. m n m n Product rule m 5. n m n Quotient rule 6. (m)n mn Power of power rule 7. (b)n nbn Power of product rule n n 8. n Power of quotient rule b b helpful hint Applictions In this section we use the mount formul for interest compounded nnully only. But you probbly hve money in bnk where interest is compounded dily. In this cse r represents the dily rte (APR 65) nd n is the number of dys tht the money is on deposit. Amount Formul The mount A of n investment of P dollrs with interest rte r compounded nnully for n yers is given by the formul A P( r)n. 6 Finding the mount According to Fidelity Investments of Boston, U.S. common stocks hve returned n verge of 0% nnully since 96. If your gret-grndfther hd invested $00 in the stock mrket in 96 nd obtined the verge increse ech yer, then how much would the investment be worth in the yer 006 fter 80 yers of growth? Both positive nd negtive eponents occur in formuls used in investment situtions. The mount of money invested is the principl, nd the vlue of the principl fter certin time period is the mount. Interest rtes re nnul percentge rtes.

16 7 (5-6) Chpter 5 Eponents nd Polynomils Use n 80, P $00, nd r 0.0 in the mount formul: A P( r)n A 00( 0.0)80 00(.)80 0,80.0 So $00 invested in 96 would hve mounted to $0,80.0 in 006. clcultor close-up With grphing clcultor you cn enter 00( 0.0)80 lmost s it ppers in print. When we re interested in the principl tht must be invested tody to grow to certin mount, the principl is clled the present vlue of the investment. We cn find formul for present vlue by solving the mount formul for P : A P( r)n A P n ( r) P A( r) n Divide ech side by ( r)n. Definition of negtive eponent Present Vlue Formul The present vlue P tht will mount to A dollrs fter n yers with interest compounded nnully t nnul interest rte r is given by P A( r) n. 7 Finding the present vlue If your gret-grndfther wnted you to hve $,000,000 in 006, then how much could he hve invested in the stock mrket in 96 to chieve this gol? Assume he could get the verge nnul return of 0% (from Emple 6) for 80 yers. Use r 0.0, n 80, nd A,000,000 in the present vlue formul: P P P P A( r) n,000,000( 0.0) 80,000,000(.) 80 Use clcultor with n eponent key A deposit of $88.9 in 96 would hve grown to $,000,000 in 80 yers t rte of 0% compounded nnully. WARM-UPS True or flse? Eplin your nswer. Assume ll vribles represent nonzero rel numbers.. () 5 Flse. () 7 Flse True. ( ) 8 True 5. () 6 Flse. ( ) 9 True 6. ( y) 9y9 Flse True True 6 True

17 5. 5. (5-7) The Power Rules 7 EXERCISES Reding nd Writing After reding this section, write out the nswers to these questions. Use complete sentences.. Wht is the power of power rule? Simplify. See Emple. w y 7. y. Wht is the power of product rule?. Wht is the power of quotient rule?. Wht is principl? Principl is the mount of money invested initilly. For ll eercises in this section, ssume the vribles represent nonzero rel numbers nd use positive eponents only in your nswers. Use the rules of eponents to simplify ech epression. See Emple. 7. () 8. () 9. ( y)5. (). (m ) 6 6. (m ) (m ) 5. ( )( ) ( ) 7. ( ) 5 () 8. ( ). ( 5w ). ( w 5) y ( y ). ( b 5. (b ) 6. ( y) ) y 59. z z 60. y y ( ) b 8. (5b) (y) 0. 8 y y 7. ( y) (b) 9. 8 b. b. c b 6. b ( ) 7 Use the rules of eponents to simplify ech epression. y y y y (5 b) (mn ) (5b ) mn5 0. ( ) Use the rules of eponents to simplify ech epression. If possible, write down only the nswer () 57. ( ) 6. Simplify. See Emple. 9. ( 9y) Simplify ech epression. Assume tht the vribles represent integers. See Emple t 5t 8. n n w w 9. ( ) (6) m 6 7 p 5. m p 5. 8 (8 ) 5. (5 y)(5y ). ( )7. ( ) 6. Wht formul is used for computing the present vlue of n mount in the future with interest compounded nnully? Simplify. See Emple y 5. Wht formul is used for computing the mount of n investment for which interest is compounded nnully? 0. (6) 5. b b 6. y 8. m. 5

18 7 (5-8) Chpter 5 ( y) 7. (y 7) (y ) 6 b 7. c Eponents nd Polynomils ( y) 7. (9 9y5) (y ) 7y 7. (7z) z ( b) Write ech epression s rised to power. Assume tht the vribles represent integers m m 80. 6n 5 5 n 79. n b) How much more would your $0,000 investment be worth in 000 if you hd invested in stocks? 9. Sving for college. Mr. Wtkins wnts to hve $0,000 in svings ccount when his little Wnd is redy for college. How much must he deposit tody in n ccount pying 7% compounded nnully to hve $0,000 in 8 yers? $, Sving for retirement. In the 990s returns on Tresury Bills fell to n verge of.5% per yer (Fidelity Investments). Wilm wnts to hve $,000,000 when she retires in 5 yers. If she ssumes n verge nnul return of.5%, then how much must she invest now in Tresury Bills to chieve her gol? $75, Life epectncy of white mles. Strnge s it my seem, your life epectncy increses s you get older. The function n 8. 8n L 7.(.00) Use clcultor to evlute ech epression. Round pproimte nswers to three deciml plces. (.5) 8. 5 (.5) (0.06) (.9) 88. (.7) 5(0.7) (5.7) (.9) 90. [5.9 (0.7) ] (.76) cn be used to model life epectncy L for U.S. white mles with present ge (Ntionl Center for Helth Sttistics, ) To wht ge cn 0-yer-old white mle epect to live? 75. yers b) To wht ge cn 60-yer-old white mle epect to live? (See lso Chpter Review Eercises 5 nd 5.) 8. yers 96. Life epectncy of white femles. Life epectncy improved more for femles thn for mles during the 90s nd 950s due to drmtic decrese in mternl mortlity rtes. The function L 78.5(.00) cn be used to model life epectncy L for U.S. white femles with present ge. ) To wht ge cn 0-yer-old white femle epect to live? 80. yers b) Bob, 0, nd Ashley, 6, re n verge white couple. How mny yers cn Ashley epect to live s widow? 7.9 yers c) Why do the life epectncy curves intersect in the ccompnying figure? At 80 both mles nd femles cn epect bout 5 more yers Life epectncy (yers) 50 Stocks 50 Bonds 0 White femles White mles 0 60 Present ge 80 FIGURE FOR EXERCISES 95 AND 96 FIGURE FOR EXERCISE Number of yers fter 990 Vlue of $0,000 investment (in thousnds of dollrs) Solve ech problem. See Emples 6 nd Deeper in debt. Meliss borrowed $0,000 t % compounded nnully nd mde no pyments for yers. How much did she owe the bnk t the end of the yers? (Use the compound interest formul.) $56, Compring stocks nd bonds. According to Fidelity Investments of Boston, throughout the 990s nnul returns on common stocks verged 9%, wheres nnul returns on bonds verged 9%. ) If you hd invested $0,000 in bonds in 990 nd chieved the verge return, then wht would your investment be worth fter 0 yers in 000? $,67.6

19 5. Addition, Subtrction, nd Multipliction of Polynomils 97. Discussion. For which vlues of nd b is it true tht (b) b? Find pir of nonzero vlues for nd b for which ( b) b. 98. Writing. Eplin how to evlute in three differ ent wys. 99. Discussion. Which of the following epressions hs vlue different from the others? Eplin. ) b) 0 c) d) ( ) e) ( ) d 00. True or Flse? Eplin your nswer. ) The squre of product is the product of the squres. b) The squre of sum is the sum of the squres. ) True b) Flse 0. The function y 7.(.00) gives the life epectncy y of U.S. white mle with present ge. (See Eercise 95.) ) Grph y 7.(.00) nd y 86 on grphing clcultor. Use viewing window tht shows the intersection of the two grphs. b) Use the intersect feture of your clcultor to find the point of intersection. c) Wht does the -coordinte of the point of intersection tell you? b) (87.5, 86) c) At 87.5 yers of ge you cn epect to live until 86. The model fils here. G R A P H I N G C ALC U L ATO R EXERCISES 0. At % compounded nnully the vlue of n investment of $0,000 fter yers is given by y 0,000(.). In this section Polynomils Evluting Polynomils Addition nd Subtrction of Polynomils ADDITION, SUBTRACTION, AND MULTIPLICATION OF POLYNOMIALS A polynomil is prticulr type of lgebric epression tht serves s fundmentl building block in lgebr. We used polynomils in Chpters nd, but we did not identify them s polynomils. In this section you will lern to recognize polynomils nd to dd, subtrct, nd multiply them. Polynomils The epression 5 7 is n emple of polynomil in one vrible. Becuse this epression could be written s Multipliction of Polynomils ( 5 ) 7 ( ), we sy tht this polynomil is sum of four terms:, 5, 7, nd. A term of polynomil is single number or the product of number nd one or more vribles rised to whole number powers. The number preceding the vrible in ech term is clled the coefficient of tht vrible. In 5 7 the coefficient of is, the coefficient of is 5, nd the coefficient of is 7. In lgebr number is frequently referred to s constnt, nd so the lst term is clled the constnt term. A polynomil is defined s single term or sum of finite number of terms. 75 ) Grph y 0,000(.) nd the function y 0,000 on grphing clcultor. Use viewing window tht shows the intersection of the two grphs. b) Use the intersect feture of your clcultor to find the point of intersection. c) The -coordinte of the point of intersection is the number of yers tht it will tke for the $0,000 investment to double. Wht is tht number of yers? GET TING MORE INVOLVED 5. (5-9)

20 76 (5-0) Chpter 5 Eponents nd Polynomils Identifying polynomils Determine whether ech lgebric epression is polynomil. c) y ) b) d) e) 9 8 ) The number is polynomil of one term, constnt term. b) Since cn be written s, it is polynomil of two terms. c) The epression y is not polynomil becuse hs negtive eponent. d) If this epression is rewritten s, then it fils to be polynomil becuse of the negtive eponents. So polynomil does not hve vribles in denomintors, nd is not polynomil. e) The epression 9 8 is polynomil. For simplicity we usully write polynomils in one vrible with the eponents in decresing order from left to right. Thus we would write 5 7 rther thn 5 7. When polynomil is written in decresing order, the coefficient of the first term is clled the leding coefficient. Certin polynomils hve specil nmes depending on the number of terms. A monomil is polynomil tht hs one term, binomil is polynomil tht hs two terms, nd trinomil is polynomil tht hs three terms. The degree of polynomil in one vrible is the highest power of the vrible in the polynomil. The number 0 is considered to be monomil without degree becuse 0 0 n, where n could be ny number. study Identifying coefficients nd degree Stte the degree of ech polynomil nd the coefficient of. Determine whether the polynomil is monomil, binomil, or trinomil. b) 8 c) 6 ) 5 7 ) The degree of this trinomil is, nd the coefficient of is. b) The degree of this binomil is 8, nd the coefficient of is. c) Becuse 6 6 0, the number 6 is monomil with degree 0. Becuse does not pper in this polynomil, the coefficient of is 0. tip Although we re minly concerned here with polynomils in one vrible, we will lso encounter polynomils in more thn one vrible, such s 5y 6y, Effective time mngement will llow dequte time for school, socil life, nd free time. However, t times you will hve to scrifice to do well. y z, nd b c.

21 5. Addition, Subtrction, nd Multipliction of Polynomils (5-) 77 In term contining more thn one vrible, the coefficient of vrible consists of ll other numbers nd vribles in the term. For emple, the coefficient of in 5y is 5y, nd the coefficient of y is 5. Evluting Polynomil Functions study The formul D 6t v0t s0 is used to model the effect of grvity on n object tossed stright upwrd with initil velocity v0 feet per second from n initil height of s0 feet. For emple, if bll is tossed into the ir t 6 feet per second from height of feet, then D 6t 6t gives the bll s distnce bove the ground in feet, t seconds fter it is tossed. Becuse D is determined by t, we sy tht D is function of t. The vlues of t rnge from t 0 when the bll is tossed to the time when it hits the ground. To emphsize tht the vlue of D depends on t, we cn use the function nottion introduced in Chpter nd write tip A lumber mill turns logs into plywood, dding vlue to the logs. College is like lumber mill. If you re not chnging, growing, nd lerning, you my not be incresing in vlue. Everything tht you lern increses your vlue. D(t) 6t 6t. We red D(t) s D of t. The epression D(t) is the vlue of the polynomil t time t. To find the vlue when t, replce t by : D() The sttement D() 68 mens tht the bll is 68 feet bove the ground seconds fter the bll ws tossed upwrd. Note tht D() does not men D times. Finding the vlue of polynomil Suppose Q() 7 6. Find Q() nd Q( ). To find Q(), replce by in Q() 7 6: Q() To find Q( ), replce by in Q() 7 6: Q( ) ( ) ( ) 7( ) So Q() 0 nd Q( ). Addition nd Subtrction of Polynomils When evluting polynomil, we get rel number. So the opertions tht we perform with rel numbers cn be performed with polynomils. Actully, we hve been dding nd subtrcting polynomils since Chpter. To dd two polynomils, we simply dd the like terms.

22 78 (5-) Chpter 5 helpful Eponents nd Polynomils Adding polynomils Find the sums. ) ( 5 7) (7 0) b) ( 5 7) ( ) ) ( 5 7) (7 0) 8 9 Combine like terms. b) For illustrtion we will write this ddition verticlly: 7 5 Line up like terms. Add. hint When we perform opertions with polynomils nd write the results s equtions, those equtions re identities. For emple, ( ) ( 7) 5 8 When we subtrct polynomils, we subtrct like terms. Becuse b ( b), we often perform subtrction by chnging signs nd dding. is n identity. helpful 5 Subtrcting polynomils Find the differences. ) ( 7 ) (5 6 ) b) (6y z 5yz 7) (y z yz 9) ) We find the first difference horizontlly: hint ( 7 ) (5 6 ) For subtrction, write the originl problem nd then rewrite it s ddition with the signs chnged. Mny students hve trouble when they write the originl problem nd then overwrite the signs. Verticl subtrction is essentil for performing long division of polynomils in Section 5.5. Chnge signs. Combine like terms. b) For illustrtion we write (6yz 5yz 7) (yz yz 9) verticlly: 5yz 7 6y z y z yz 9 Chnge signs. 6y z yz yz 6 Add. It is certinly not necessry to write out ll of the steps shown in Emples nd 5, but we must use the following rule. Addition nd Subtrction of Polynomils To dd two polynomils, dd the like terms. To subtrct two polynomils, subtrct the like terms. Multipliction of Polynomils We lerned how to multiply monomils when we lerned the product rule in Section 5.. For emple, 8 5. To multiply monomil nd polynomil of two or more terms, we pply the distributive property. For emple, ( 5) 5.

23 5. 6 Addition, Subtrction, nd Multipliction of Polynomils Multiplying monomil nd polynomil Find the products. b) ( )(5 ) ) b b (5-) 79 c) ( 5 )( ) ) b b 6b b) ( )(5 ) 5 5 c) Ech term of 5 is multiplied by : ( 5 )( ) 5 6 Note wht hppened to the binomil in Emple 6(b) when we multiplied it by. If we multiply ny difference by, we get the sme type of result: ( b) b b. Becuse multiplying by is the sme s tking the opposite, we cn write this eqution s ( b) b. This eqution sys tht b nd b re opposites or dditive inverses of ech other. Note tht the opposite of b is b, not b. To multiply binomil nd trinomil, we cn use the distributive property or set it up like multipliction of whole numbers. helpful 7 Multiplying binomil nd trinomil Find the product ( )( 5). We cn find this product by pplying the distributive property twice. First we multiply the binomil nd ech term of the trinomil: hint ( )( 5) ( ) ( ) ( )( 5) Mny students find verticl multipliction esier thn pplying the distributive property twice horizontlly. However, you should lern both methods becuse horizontl multipliction will help you with fctoring by grouping in Section Distributive property Distributive property Combine like terms. We could hve found this product verticlly: ( 5) 6 0 ( 5) 5 Add. Multipliction of Polynomils To multiply polynomils, multiply ech term of the first polynomil by ech term of the second polynomil nd then combine like terms. In the net emple we multiply binomils.

24 80 (5-) Chpter 5 Eponents nd Polynomils 8 Multiplying binomils Find the products. ) ( y)(z ) b) ( )( 5) ) ( y)(z ) ( y)z ( y) Distributive property z yz y Distributive property Notice tht this product does not hve ny like terms to combine. b) Multiply: WARM-UPS True or flse? Eplin your nswers The epression 5 is trinomil. Flse In the polynomil 5 the coefficient of is 5. Flse The degree of the polynomil 5 is. Flse If C(), then C(5). True If P(t) 0t 0, then P(0) 0. Flse ( 5) ( 5 7) for ny vlue of. True ( 5) ( ) 8 for ny vlue of. Flse ( ) 8 6 for ny vlue of. True ( 7) 7 for ny vlue of. True The opposite of y 5 is y 5 for ny vlue of y. Flse EXERCISES Reding nd Writing After reding this section, write out the nswers to these questions. Use complete sentences. 6. Wht property is used when multiplying binomil nd trinomil?. Wht is term of polynomil? A term of polynomil is single number or the product of Determine whether ech lgebric epression is polynomil. See Emple. 7. Yes 8. 9 Yes 9. No No. y y Stte the degree of ech polynomil nd the coefficient of. Determine whether ech polynomil is monomil, binomil, or trinomil. See Emple.. Wht is coefficient?. Wht is constnt?. Wht is polynomil? 5. Wht is the degree of polynomil? 5. 8

25 dugoi_c05.qd_a 0/5/00 0:8 AM Pge 8 5. Addition, Subtrction, nd Multipliction of Polynomils ( )( ) (5-5) 8 5. ( )( ) 5. ( )( ) 5. ( )( ) 8 Find ech product verticlly. See Emples Multiply 56. Multiply Multiply 58. Multiply b 5 5 b 0 5 b 59. Multiply 60. Multiply Multiply 6. Multiply b b y y y b b y For ech given polynomil, find the indicted vlue of the polynomil. See Emple.. P(), P(). P(), P( ) 5. M() 9, M( ) 6. C(w) w w, C(0) 7. R() 5, R() 8. T() 7 6, T( ) Perform the indicted opertions. See Emples nd ( ) ( 5) 0. (w 6) (w 5). (7y 0) (y 5) 5. (5b 7) (b 6). ( ) ( 5 9). (y y 8) (y y ) 5. ( ) ( 5) 6. ( 5) ( ) Perform the indicted opertions. 6. ( 7) ( ) (5 ) 5 6. (5 ) ( ) ( ) 65. ( 5 ) ( 6 7) 66. (w w ) (w w) 67. (w 7w ) (w w 5) 68. ( ) ( ) 69. ( )( ) 70. ( )( 9) 7. ( w )(z w) 7. (w )(t ) w 7. (y )(y 5) 7. (b )(b 8) 8 Perform the indicted opertions verticlly. See Emples nd Add 8. Add Subtrct 0. Subtrct Subtrct. Subtrct Add. Add y w y w w Find ech product. See Emples ( b5)( b) 5 7. ( ) 8. ( 9) y(y ) 50. yz(8yz yz y) 6 7 Perform the following opertions using clcultor. 75. (. 5.)(6.5.8) ( 0.8)(.6.) ( ) ( ) 78. ( ) ( ).9 5. Perform the indicted opertions

26 (5-6) Chpter 5 6 Eponents nd Polynomils 0. Mrginl profit. A compny uses the formul P(n) n 0.9n to estimte its dily profit in dollrs for producing n utomtic grge door openers. The mrginl profit of the nth opener is the mount of dditionl profit mde for tht opener. For emple, the mrginl profit for the fourth opener is P() P(). Find the mrginl profit for the fourth opener. Wht is the mrginl profit for the tenth opener? Use the br grph to eplin why the mrginl profit increses s production goes up. $7.0, $7.90. There is greter difference in height between djcent brs s the number of openers increses [ ( 5 )] [ ( 5)] 8. [ ( ) 5] [ 5( ) ] [5 ( )][ 7( )] 86. [ (5 )][ (5 )] 87. [ (m )][ (m )] [ ( )][ ( )] Profit (in dollrs) 89. (5 ) [5 ( 7)] ( ) 5[ ( 6)] Number of grge door openers Perform the indicted opertions. A vrible used in n eponent represents n integer; vrible used s bse represents nonzero rel number. 9. (m m ) ( 5m 7m 8) 9. (bz 6) (bz bz 7) 9. ( n )( n ) FIGURE FOR EXERCISE 0 0. Mle nd femle life epectncy. Since 950 the life epectncies of U.S. mles nd femles born in yer y cn be modeled by the formuls M( y) 0.65y 5.9 nd F(y) 0.868y 8.98, respectively (Ntionl Center for Helth Sttistics, ) How much greter ws the life epectncy of femle born in 950 thn mle born in 950? b) Are the lines in the ccompnying figure prllel? c) In wht yer will femle life epectncy be 8 yers greter thn mle life epectncy? ) 6. yers b) no c) More life epectncy. Use the functions from the previous eercise for these questions. ) A mle born in 975 does not wnt his future wife to outlive him. Wht should be the yer of birth for his 9. (y )(y 7) 8y t t 95. zw zw(z w zw) 5z 96. (w p )(w p w p ) 97. ( r y)(r ry y) 98. ( z)( z) Solve ech problem. See Eercises Cost of grvel. The cost in dollrs of cubic yrds of grvel is given by the formul C() 0 5. Find C(), the cost of cubic yrds of grvel. 00. Annul bonus. Sheil s nnul bonus in dollrs for selling n life insurnce policies is given by the formul B(n) 0.n n U.S. fem U.S. les mle s Mrginl cost. A compny uses the formul C(n) 50n 0.0n to find the dily cost in dollrs of mnufcturing n luminum windows. The mrginl cost of the nth window is the dditionl cost incurred for mnufcturing tht window. For emple, the mrginl cost of the third window is C() C(). Find the mrginl cost for mnufcturing the third window. Wht is the mrginl cost for mnufcturing the tenth window? $9.5, $5.6 y Life epectncy (yers) Find B(0), her bonus for selling 0 policies. $50 Yer of birth FIGURE FOR EXERCISES 0 AND 0

27 5. wife so tht they both cn be epected to die in the sme yer? M(y) F( y) b) Find to get formul for the life e pectncy of person born in yer y. ) 969 b) 0.76y 68.5 (5-7) Discussion. Give n emple of two fourth-degree trinomils whose sum is third-degree binomil. 07. Coopertive lerning. Work in group to find the product ( b)(c d). How mny terms does it hve? Find the product ( b)(c d )(e f ). How mny terms does it hve? How mny terms re there in product of four binomils in which there re no like terms to combine? How mny terms re there in product of n binomils in which there re no like terms? GET TING MORE INVOLVED 05. Discussion. Is it possible for binomil to hve degree? If so, give n emple. 5. Multiplying Binomils MULTIPLYING BINOMIALS In Section 5. you lerned to multiply polynomils. In this section you will lern rules to mke multipliction of binomils simpler. In this section The FOIL Method The FOIL Method The Squre of Binomil Product of Sum nd Difference helpful Consider how we find the product of two binomils nd 5 using the distributive property twice: ( )( 5) ( ) ( ) Distributive property Distributive property Combine like terms. There re four terms in the product. The term is the product of the first term of ech binomil. The term 5 is the product of the two outer terms, 5 nd. The term is the product of the two inner terms, nd. The term 5 is the product of the lst two terms in ech binomil, nd 5. It my be helpful to connect the terms multiplied by lines. hint The product of two binomils lwys hs four terms before combining like terms. The product of two trinomils lwys hs nine terms before combining like terms. How mny terms re there in the product of binomil nd trinomil? L F ( )( 5) I O F O I L First terms Outer terms Inner terms Lst terms So insted of writing out ll of the steps in using the distributive property, we cn get the result by finding the products of the first, outer, inner, nd lst terms. This method is clled the FOIL method. For emple, let s pply FOIL to the product ( )( ): L F F O I L ( )( ) I O If the outer nd inner products re like terms, you cn sve step by writing down only their sum.

28 8 (5-8) Chpter 5 helpful Eponents nd Polynomils Multiplying binomils Use FOIL to find the products of the binomils. ) ( )( ) b) ( 5)( 5) c) (m w)(m w) d) ( b)( ) hint You my hve to prctice FOIL while to get good t it. However, the better you re t FOIL, the esier you will find fctoring in Sections 5.6, 5.7, nd 5.8. F O I L ) ( )( ) b) ( 5)( 5) c) (m w)(m w) m mw mw w m mw w d) ( b)( ) b b There re no like terms. The Squre of Binomil hint helpful To find ( b), the squre of sum, we cn use FOIL on ( b)( b): To visulize the squre of sum, drw squre with sides of length b s shown. b b b b b ( b)( b) b b b b b You cn use the result b b tht we obtined from FOIL to quickly find the squre of ny sum. To squre sum, we squre the first term (), dd twice the product of the two terms (b), then dd the squre of the lst term (b ). Rule for the Squre of Sum ( b) b b The re of the lrge squre is ( b). It comes from four terms s stted in the rule for the squre of sum. In generl, the squre of sum ( b) is not equl to the sum of the squres b. The squre of sum hs the middle term b. Squring binomil Squre ech sum, using the new rule. b) (w ) ) ( 5) c) (y ) ) ( 5) ()(5) helpful Squre of first hint You cn use Squre of lst b) (w ) (w) (w)() w w 9 c) (y ) (y) (y)() y8 y 9 ( 5) 0 5 ( 0) 5 to lern trick for squring number tht ends in 5. For emple, to find 5, find or 65. More simply, to find 5, find nd follow tht by 5: 5 5. CAUTION Squring 5 correctly, s in Emple (), gives us the identity ( 5) 0 5, which is stisfied by ny. If you forget the middle term nd write ( 5) 5, then you hve n eqution tht is stisfied only if 0. Twice the product

29 5. Multiplying Binomils (5-9) 85 To find ( b), the squre of difference, we cn use FOIL: ( b)( b) b b b b b As in squring sum, it is simply better to remember the result of using FOIL. To squre difference, squre the first term, subtrct twice the product of the two terms, nd dd the squre of the lst term. Rule for the Squre of Difference ( b) b b helpful Squring binomil Squre ech difference, using the new rule. b) (w 5y) c) ( st) ) ( 6) ) ( 6) ()(6) 6 hint d) ( 5) For the middle term, subtrct twice the product: ()(6). 6 b) (w 5y) (w) (w)(5y) (5y) 9w 0wy 5y c) ( st) ( ) ( )(st) (st) 6 8st st d) ( 5) ()(5) (5) Mny students keep using FOIL to find the squre of sum or difference. However, you will be gretly rewrded if you lern the new rules for squring sum or difference. Product of Sum nd Difference If we multiply the sum b nd the difference b by using FOIL, we get helpful ( b)( b) b b b b. hint The inner nd outer products dd up to zero, cnceling ech other out. So the product of sum nd difference is the difference of two squres, s shown in the following rule. You cn use ( b)( b) b to perform mentl rithmetic tricks such s Rule for the Product of Sum nd Difference ( b)( b) b Wht is 9 5? 8? Finding the product of sum nd difference Find the products. ) ( )( ) b) ( 8)( 8) c) ( y )( y ) ) ( )( ) 9 b) ( 8)( 8) 6 6 c) ( y)( y ) 9 y6

30 86 (5-0) Chpter 5 Eponents nd Polynomils The squre of sum, the squre of difference, nd the product of sum nd difference re referred to s specil products. Although the specil products cn be found by using the distributive property or FOIL, they occur so frequently in lgebr tht it is essentil to lern the new rules. In the net emple we use the specil product rules to multiply two trinomils nd to squre 5 Using specil product rules to multiply trinomils Find the products. ) [( y) ][( y) ] b) [(m n) 5] ) Use the rule ( b)( b) b with y nd b : [( y) ][( y) ] ( y) y y 9 b) Use the rule ( b) b b with m n nd b 5: [(m n) 5] (m n) (m n)5 5 m mn n 0m 0n 5 WARM-UPS True or flse? Eplin your nswer ( )( 5) 7 0 for ny vlue of. True ( )( 5) 6 5 for ny vlue of. True ( ) Flse ( 7) 9 for ny vlue of. True (8 ) 6 9 Flse The product of sum nd difference of the sme two terms is equl to the difference of two squres. True (60 )(60 ) 600 True ( y) y y for ny vlues of nd y. True ( ) 9 for ny vlue of. Flse The epression 5 is product of two binomils. Flse EXERCISES Reding nd Writing After reding this section, write out the nswers to these questions. Use complete sentences.. Wht property is used to multiply two binomils? 5. How do you squre difference? The squre of difference is the squre of the first term the lst term. 6. How do you find the product of sum nd difference? The product of sum nd difference of the sme two. Wht does FOIL stnd for?. Wht is the purpose of the FOIL method? find the. How do you squre sum? first term plus twice Find ech product. When possible, write down only the nswer. See Emple. 7. ( )( ) 8 8. ( )( 5) 5

31 5. 9. ( )( ) 0. (y )(y ). ( )( 5). ( 5)( 6). ( 7)( 7). (y 8)(y 8) 5. ( )( ) 6. (t )(t ) 7. (6z w)(w z) w 8. (y w)(w y) 9. (k t)(t k) 0. (7 )( ) 7. ( )(y w) y. (z )( y ) yz y z Find the squre of ech sum or difference. When possible, write down only the nswer. See Emples nd.. (m ) m. ( ) 5. ( ) 6. (b ) b 7. (w ) 8. (m ) 9. (t 5u) 0. (w ). ( ). ( d 5). ( y). (m 5n) Find ech product. See Emple. 7. (w 9)(w 9) w 8. (m )(m ) m 9. (w y)(w y) 0. ( )( ). ( 7)( 7). (5 )(5 ). ( )( ). (y )(y ) Use the specil product rules to find ech product. See Emple [(m t) 5][(m t) 5] 6. [( ) y][( ) y] 7. [y (r 5)][y (r 5)] y 8. [ ( k)][ ( k)] 9. [(y t) ] y yt 50. [(u v) ] u 6uv 5. [h (k )] 5. [p (q 6)] 87 6q 6 Perform the opertions nd simplify. 5. ( 6)( 9) 5. ( )( ) 55. (5 )(5 ) ( b)( b) 57. ( )( 5) 58. (5 )(5 ) 59. (t )(t w) t 60. (5 9)( b) 5 6. ( y) 9 6. (5 b) 5 6. (y )(y 5) 6y 6. (b )(b ) (50 )(50 ) 66. (00 )(00 ) 67. ( 7) ( pq) pq 69. y y 6y 70. 5y y 00y 5 0 w b 9b 7. ( h) h ( h) 7. h 7. ( )( ) 7. ( )( ) 75. (y ) y 9y 76. (b ) 6b (5-) Multiplying Binomils 8 8b Use clcultor to help you perform the following opertions. 77. (..5)(5..9) (5. 9.) (.6y.) y.68y (. 7.9b)(. 7.9b) 6.b Find the products. Assume ll vribles re nonzero nd vribles used in eponents represent integers. ( m )( m ) 6 ( n b)( n b) n ( n n ) b( b b 5) ( m n) ( w t ) m k m (5y 8z )(y z k) 5y y z 0y z b 5 b 88. ( y )( y ) y 8 6y

32 88 (5-) Chpter 5 Eponents nd Polynomils re then folded up, nd the corners re seled. Find polynomil tht gives the volume of the pn in cubic feet ( ft ). Wht is the volume of the pn if inches? Solve ech problem. 89. Are of room. Suppose the length of rectngulr room is meters nd the width is meters. Find trinomil tht cn be used to represent the re of the room. 90. House plns. Brbie nd Ken plnned to build squre house with re squre feet. Then they revised the pln so tht one side ws lengthened by 0 feet nd the other side ws shortened by 6 feet. Find trinomil tht cn be used to represent the re of the revised house. 0 squre feet in. ft 6 ft 6 ft FIGURE FOR EXERCISE 9 ft ft 0 ft FIGURE FOR EXERCISE Avilble hbitt. A wild niml will generlly sty more thn kilometers from the edge of forest preserve. So the vilble hbitt for the niml ecludes n re of uniform width on the edge of the rectngulr forest preserve shown in the figure. The vlue of depends on the niml. Find trinomil in tht gives the re of the vilble hbitt in squre kilometers (km) for the forest preserve shown. Wht is the vilble hbitt in this forest preserve for bobct for which 0. kilometers. Forest preserve Avilble hbitt 8 km 0 km FIGURE FOR EXERCISE 9 9. Energy efficient. A mnufcturer of mobile homes mkes custom model tht is feet long, feet wide, nd 8 feet high (ll inside dimensions). The insultion is inches thick in the wlls, 6 inches thick in the floor, nd 8 inches thick in the ceiling. Given tht the insultion costs the mnufcturer 5 cents per cubic foot nd doors nd windows tke up 80 squre feet of wll spce, find polynomil in tht gives the cost in dollrs for insultion in this model. Stte ny ssumptions tht you re mking to solve this problem..5 7 dollrs GET TING MORE INVOLVED 95. Eplortion. ) Find ( b) by multiplying ( b) by b. b) Net find ( b) nd ( b)5. c) How mny terms re in ech of these powers of b fter combining like terms? d) Mke generl sttement bout the number of terms in ( b)n. 96. Coopertive lerning. Mke four-column tble with columns for, b, ( b), nd b. Work with group to fill in the tble with five pirs of numbers for nd b for which ( b) b. For wht vlues of nd b does ( b) b? 97. Discussion. The re of the lrge squre shown in the figure is ( b). Find the re of ech of the four smller regions in the figure, nd then find the sum of those res. Wht conclusion cn you drw from these res bout ( b)? 9. Cubic coting. A cubic metl bo inches on ech side is designed for trnsporting frozen specimens. The bo is surrounded on ll sides by -inch-thick lyer of styrofom insultion. Find polynomil tht represents the totl volume of the cube nd styrofom. 9. Overflow pn. An ir conditioning contrctor mkes n overflow pn for condenser by cutting squres with side of length feet from the corners of foot by 6 foot piece of glvnized sheet metl s shown in the figure. The sides b b b b FIGURE FOR EXERCISE 97

33 Division of Polynomils (5-) 89 DIVISION OF POLYNOMIALS We begn our study of polynomils in Section 5. by lerning how to dd, subtrct, nd multiply polynomils. In this section we will study division of polynomils. In this section Dividing Polynomil by Monomil Dividing Polynomil by Binomil Synthetic Division Division nd Fctoring Dividing Polynomil by Monomil You lerned how to divide monomils in Section 5.. For emple, 6 6 (). We check by multiplying. Becuse 6, this nswer is correct. Recll tht b c if nd only if c b. We cll the dividend, b the divisor, nd c the quotient. We my lso refer to b nd s quotients. b We cn use the distributive property to find tht ( 5 ) 6 5. So if we divide 6 5 by the monomil, we must get 5. We cn perform this division by dividing into ech term of 6 5 : In this cse the divisor is, the dividend is 6 5, nd the quotient is 5. helpful Dividing polynomils Find the quotient. ) 5 ( ) ) When dividing 5 by, we subtrct the eponents: 5 5 ( ) 6 The quotient is 6. Check: 6 5 b) Divide ech term of by : The quotient is 5. Check: ( 5 ) hint Recll tht the order of opertions gives multipliction nd division n equl rnking nd sys to do them in order from left to right. So without prentheses, 5 ctully mens 5. b) ( ) ( )

34 90 (5-) Chpter 5 Eponents nd Polynomils Dividing Polynomil by Binomil We cn multiply nd 5 to get ( )( 5) 0. So if we divide 0 by the fctor, we should get the other fctor 5. This division is not done like division by monomil; it is done like long division of whole numbers. We get the first term of the quotient by dividing the first term of into the first term of 0. Divide by to get. 0 5 Multiply: ( ). Subtrct: ( ) 5. Now bring down 0. We get the second term of the quotient (below) by dividing the first term of into the first term of 5 0. Divide 5 by to get Multiply: 5( ) 5 0. Subtrct: 0 ( 0) 0. So the quotient is 5 nd the reminder is 0. If the reminder is not 0, then dividend (divisor)(quotient) (reminder). If we divide ech side of this eqution by the divisor, we get dividend reminder quotient. divisor divisor When dividing polynomils, we must write the terms of the divisor nd the dividend in descending order of the eponents. If ny terms re missing, s in the net emple, we insert terms with coefficient of 0 s plceholders. When dividing polynomils, we stop the process when the degree of the reminder is smller thn the degree of the divisor. helpful Dividing polynomils Find the quotient nd reminder for ( 5) ( ). Rerrnge 5 s 5 nd insert the terms 0 nd 0 : ( 9 ) hint Students usully hve the most difficulty with the subtrction prt of long division. So py prticulr ttention to tht step nd double check your work.

35 5.5 Division of Polynomils (5-5) 9 The quotient is 9 7, nd the reminder is 76. Note tht the degree of the reminder is, nd the degree of the divisor is. To check, verify tht ( )( 9 7) Rewriting rtio of two polynomils 9 Write in the form reminder quotient. divisor study Divide 9 by. Insert 0 for the missing term. 0 9 () ( 6 ) ( 9) ( ) Since the quotient is nd the reminder is, we hve 9. To check the nswer, we must verify tht ( )( ) 9. tip Hve you ever used ecuses to void studying? ( Before I cn study, I hve to do my lundry nd go to the bnk. ) Since the verge ttention spn for one tsk is pproimtely 0 minutes, it is better to tke breks from studying to run errnds nd to do lundry thn to get everything done before you strt studying. Synthetic Division When dividing polynomil by binomil of the form c, we cn use synthetic division to speed up the process. For synthetic division we write only the essentil prts of ordinry division. For emple, to divide 5 by, we write only the coefficients of the dividend, 5,, nd in order of descending eponents. From the divisor we use nd strt with the following rrngement: 5 ( 5 ) ( ) Net we bring the first coefficient,, stright down: 5 Bring down We then multiply the by the from the divisor, plce the nswer under the 5, nd then dd tht column. Using for llows us to dd the column rther thn subtrct s in ordinry division: 5 Add Multiply

36 9 (5-6) Chpter 5 Eponents nd Polynomils We then repet the multiply-nd-dd step for ech of the remining columns: Multiply Reminder Quotient From the bottom row we cn red the quotient nd reminder. Since the degree of the quotient is one less thn the degree of the dividend, the quotient is. The reminder is 7. The strtegy for getting the quotient Q() nd reminder R by synthetic division cn be stted s follows. Strtegy for Using Synthetic Division List the coefficients of the polynomil (the dividend). Be sure to include zeros for ny missing terms in the dividend. For dividing by c, plce c to the left. Bring the first coefficient down. Multiply by c nd dd for ech column. Red Q() nd R from the bottom row. CAUTION Synthetic division is used only for dividing polynomil by the binomil c, where c is constnt. If the binomil is 7, then c 7. For the binomil 7 we hve 7 ( 7) nd c 7. Using synthetic division Find the quotient nd reminder when is divided by. Since ( ), we use for the divisor. Becuse is missing in the dividend, use zero for the coefficient of : Add Multiply 0 9 Quotient nd reminder Becuse the degree of the dividend is, the degree of the quotient is. The quotient is, nd the reminder is 9. We cn lso epress the results of this minder division in the form quotient re : divisor Division nd Fctoring To fctor polynomil mens to write it s product of two or more simpler polynomils. If we divide two polynomils nd get 0 reminder, then we cn write dividend (divisor)(quotient)

37 5.5 Division of Polynomils (5-7) 9 nd we hve fctored the dividend. The dividend fctors s the divisor times the quotient if nd only if the reminder is 0. We cn use division to help us discover fctors of polynomils. To use this ide, however, we must know fctor or possible fctor to use s the divisor. 5 Using synthetic division to determine fctors Is fctor of 6 5? We cn use synthetic division to divide 6 5 by : Becuse the reminder is 0, is fctor, nd 6 5 ( )(6 6 Using division to determine fctors Is b fctor of b? Divide b by b. Insert zeros for the missing b- nd b -terms. b b b 0 0 b b b 0 b b b b b b 0 Becuse the reminder is 0, b is fctor, nd b ( b)( b b). True or flse? Eplin your nswer. WARM-UPS If b c, then c is the dividend. Flse The quotient times the dividend plus the reminder equls the divisor. Flse ( )( ) 5 7 is true for ny vlue of. True The quotient of ( 5 7) ( ) is. True If 5 7 is divided by, the reminder is. True To divide by, we use in synthetic division. Flse We cn use synthetic division to divide 6 by 5. Flse If 5 is divided by, the quotient hs degree. True If the reminder is zero, then the divisor is fctor of the dividend. True If the reminder is zero, then the quotient is fctor of the dividend. True

38 (5-8) Chpter 5 Eponents nd Polynomils EXERCISES Reding nd Writing After reding this section, write out the nswers to these questions. Use complete sentences.. Wht re the dividend, divisor, nd quotient?, nd. In wht form should polynomils be written for long division? eponents in descending order.. Wht do you do bout missing terms when dividing polynomils? (w w ) (w ) w w 6, 9 ( ) ( ), 6 ( ) ( ) 7, 8 ( ) ( ) 6, ( 6 ) ( ) 6 9,. (5 ) ( ),. ( ) ( ), ( ) ( ). ( 5 ) ( 6). When do you stop the long division process for dividing polynomils? Stop the long division process when the reminder hs 5. Wht is synthetic division used for? 6. Wht is the reltionship between division of polynomils nd fctoring polynomils? Find the quotient. See Emple ( ) 8. 0 ( 5) 9. 6 ( 8 ) 0. (). (6b 9) b. (8 6). ( 6) (). (5 0 0) (5) 5. (0 8 6 ) ( ) 6. ( 9 6 ) ( ) 7. (7 ) () 8. (6 5 ) ( ) Find the quotient nd reminder s in Emple. Check by using the formul dividend (divisor)(quotient) reminder. 9. ( 8 ) ( ) 5, 0. ( 5 7) ( ),. ( ) ( ), 8. () ( ),. ( 8) ( ), 0. (y ) (y ) y y, 0 5. ( 5) ( ) 8, 5. ( 6) ( ) 6. ( ) ( ) Write ech epression in the form reminder quotient. divisor See Emple Use synthetic division to find the quotient nd reminder when the first polynomil is divided by the second. See Emple ,, 6 5, 9, 67

39 , 6, 5. 7,, , 9, , 5 0, ,, , 5, ,, ,, , , , , 6.8 For ech pir of polynomils, determine whether the first polynomil is fctor of the second. Use synthetic division when possible. See Emples 5 nd 6. 6., 8 No 6., 8 No 6., Yes 6., 5 No 65., 6 Yes 66. 5, No 67. w, 7w Yes 68. w, 8w 7 Yes 69. 5, 5 Yes 70., 6 6 Yes 7., 6 Yes 7., 6 Yes Fctor ech polynomil given tht the binomil following ech polynomil is fctor of the polynomil , ( )( ) 7. 0, 8 ( 8)( 5) 75. w 7, w (w )(w w 9) 76. w 5, w 5 (w 5)(w 5w 5) 77. 6, ( )( ) , ( 7)( ) 79. z 6z 9, z (z )(z ) , 5 ( 5)( 5) 8. 6y 5y, y (y )(y ) 8. y y 6, y (y )(y ) Solve ech problem. 8. Averge cost. The totl cost in dollrs for mnufcturing professionl rcing bicycles in one week is given by the polynomil function Cost (in thousnds of dollrs) C() AC() Number of bicycles 5 FIGURE FOR EXERCISE 8 8. Averge profit. The weekly profit in dollrs for mnufcturing bicycles is given by the polynomil P() 00. The verge profit per bicycle is given by AP() P(). Find AP(). Find the verge profit per bicycle when bicycles re mnufctured. AP() 00, $ 85. Are of poster. The re of rectngulr poster dvertising Perl Jm concert is squre feet. If the length is feet, then wht is the width? feet 86. Volume of bo. The volume of shipping crte is h 5h 6h. If the height is h nd the length is h, then wht is the width? h h? h FIGURE FOR EXERCISE Volume of pyrmid. Ancient Egyptin pyrmid builders knew tht the volume of the truncted pyrmid shown in the figure on the net pge is given by H( b) V, ( b) C() The verge cost per bicycle is given by C() AC(). ) Find formul for AC(). AC() b) Is AC() constnt function? No c) Why does the verge cost look constnt in the ccompnying figure? Becuse AC() is very close to 00 for less thn 5, the grph looks horizontl. (5-9) Division of Polynomils

40 96 (5-0) Chpter 5 Eponents nd Polynomils where is the re of the squre bse, b is the re of the squre top, nd H is the distnce from the bse to the top. Find the volume of truncted pyrmid tht hs bse of 900 squre meters, top of 00 squre meters, nd height H of 0 meters. 6,. cubic meters 88. Egyptin pyrmid formul. Rewrite the formul of the previous eercise so tht the denomintor contins the number only. H( b b) V GET TING MORE INVOLVED b 89. Discussion. On test student divided 5 7 by nd got quotient of nd reminder 9 7. Verify tht the divisor times the quotient plus the reminder is equl to the dividend. Why ws the student s nswer incorrect? 90. Eplortion. Use synthetic division to find the quotient when 5 is divided by nd the quotient when 6 is divided by. Observe the pttern in the first two quotients nd then write the quotient for 9 divided by without dividing. b H FIGURE FOR EXERCISE In Section 5.5 you lerned tht polynomil could be fctored by using division: If we know one fctor of polynomil, then we cn use it s divisor to obtin the other fctor, the quotient. However, this technique is not very prcticl becuse the division process cn be somewht tedious, nd it is not esy to obtin fctor to use s the divisor. In this section nd the net two sections we will develop better techniques for fctoring polynomils. These techniques will be used for solving equtions nd problems in the lst section of this chpter. In this section Fctoring Out the Gretest Common Fctor (GCF) Fctoring Out the Opposite of the GCF Fctoring the Difference of Two Squres Fctoring Perfect Squre Trinomils Fctoring Difference or Sum of Two Cubes Fctoring Polynomil Completely Fctoring by Substitution FACTORING POLYNOMIALS Fctoring Out the Gretest Common Fctor (GCF) A nturl number lrger thn tht hs no fctors other thn itself nd is clled prime number. The numbers,, 5, 7,,, 7, 9, re the first nine prime numbers. There re infinitely mny prime numbers. To fctor nturl number completely mens to write it s product of prime numbers. In fctoring we might write. However, is not fctored completely s becuse is not prime. To fctor completely, we write (or ). We use the distributive property to multiply monomil nd binomil: 6( ) 6 If we strt with 6, we cn use the distributive property to get 6 6( ). We hve fctored out 6, which is common fctor of nd 6. We could hve fctored out just to get 6 ( ), but this would not be fctoring out the gretest common fctor. The gretest common fctor (GCF) is monomil tht includes every number or vrible tht is fctor of ll of the terms of the polynomil.

41 5.6 Fctoring Polynomils (5-) 97 We cn use the following strtegy for finding the gretest common fctor of group of terms. Strtegy for Finding the Gretest Common Fctor (GCF). Fctor ech term completely.. Write product using ech fctor tht is common to ll of the terms.. On ech of these fctors, use n eponent equl to the smllest eponent tht ppers on tht fctor in ny of the terms. The gretest common fctor Find the gretest common fctor (GCF) for ech group of terms. b) 0, 5b, 75b ) 8 y, 0y ) First fctor ech term completely: 8 y y 0y 5y The fctors common to both terms re,, nd y. In the GCF we use the smllest eponent tht ppers on ech fctor in either of the terms. So the GCF is y or y. b) First fctor ech term completely: 0 5 5b 5b 75b 5b The GCF is 5 or 5. To fctor out the GCF from polynomil, find the GCF for the terms, then use the distributive property to fctor it out. study Fctoring out the gretest common fctor Fctor ech polynomil by fctoring out the GCF. b) 8y 0 y ) c) ) First fctor ech term completely: 0 5, , The GCF of the three terms is 5. Now fctor 5 out of ech term: ( ) b) The GCF for 8y nd 0 y is y: 8y 0 y y(y 5) c) First fctor ech coefficient in :, , The GCF of the three terms is or : (5 ) tip Everyone hs different ttention spn. Strt by studying 0 to 5 minutes t time nd then build up to longer periods over time. In your senior yer you should be ble to concentrte on one tsk for 0 to 5 minutes without brek. Be relistic. When you cnnot remember wht you hve red nd cn no longer concentrte, tke brek. In the net emple the common fctor in ech term is binomil.

42 98 (5-) Chpter 5 Eponents nd Polynomils Fctoring out binomil Fctor. ) ( )w ( ) b) ( 9) ( 9) ) We tret like common monomil when fctoring: ( )w ( ) ( )(w ) b) Fctor out the common binomil 9: ( 9) ( 9) ( )( 9) Fctoring Out the Opposite of the GCF The GCF, the gretest common fctor, for 6 is, but we cn fctor out either or its opposite, : 6 ( ) ( ) In Emple 8 of this section it will be necessry to fctor out the opposite of the GCF. Fctoring out the opposite of the GCF Fctor out the GCF, then fctor out the opposite of the GCF. c) 5 ) 5 5y b) ) 5 5y 5( y) Fctor out 5. 5( y) Fctor out 5. b) ( ) The GCF is. ( ) Fctor out. c) 5 ( 5) ( 5) Fctor out. Fctor out. Fctoring the Difference of Two Squres A first-degree polynomil in one vrible, such s 5, is clled liner polynomil. (The eqution 5 0 is liner eqution.) Liner Polynomil helpful If nd b re rel numbers with 0, then b is clled liner polynomil. hint A second-degree polynomil such s 5 6 is clled qudrtic polynomil. Qudrtic Polynomil If, b, nd c re rel numbers with 0, then b c is clled qudrtic polynomil. The prefi qud mens four. So why is polynomil of three terms clled qudrtic? Perhps it is becuse qudrtic polynomil cn often be fctored into product of two binomils.

43 5.6 Fctoring Polynomils (5-) 99 One of the min gols of this chpter is to write qudrtic polynomil (when possible) s product of liner fctors. Consider the qudrtic polynomil 5. We recognize tht 5 is difference of two squres, 5. We recll tht the product of sum nd difference is difference of two squres: ( b)( b) b. If we reverse this specil product rule, we get rule for fctoring the difference of two squres. Fctoring the Difference of Two Squres b ( b)( b) The difference of two squres fctors s the product of sum nd difference. To fctor 5, we replce by nd b by 5 to get 5 ( 5)( 5). This eqution epresses qudrtic polynomil s product of two liner fctors. helpful 5 Fctoring the difference of two squres Fctor ech polynomil. b) 9 ) y 6 c) y Ech of these binomils is difference of two squres. Ech binomil fctors into product of sum nd difference. ) y 6 ( y 6)( y 6) We could lso write ( y 6)(y 6) becuse hint Using the power of power rule, we cn see tht ny even power is perfect squre: the fctors cn be written in ny order. b) 9 ( )( ) c) y ( y)( y) n ( n ) Fctoring Perfect Squre Trinomils The trinomil tht results from squring binomil is clled perfect squre trinomil. We cn reverse the rules from Section 5. for the squre of sum or difference to get rules for fctoring. Fctoring Perfect Squre Trinomils b b ( b) b b ( b) Consider the polynomil 6 9. If we recognize tht 6 9, then we cn see tht it is perfect squre trinomil. It fits the rule if nd b : 6 9 ( ) Perfect squre trinomils cn be identified by using the following strtegy.

44 00 (5-) Chpter 5 Eponents nd Polynomils Strtegy for Identifying Perfect Squre Trinomils A trinomil is perfect squre trinomil if. the first nd lst terms re of the form nd b,. the middle term is or times the product of nd b. We use this strtegy in the net emple. 6 Fctoring perfect squre trinomils Fctor ech polynomil. b) 9 ) 8 6 c) 9 ) Becuse the first term is, the lst is, nd ()() is equl to the middle term 8, the trinomil 8 6 is perfect squre trinomil: 8 6 ( ) b) Becuse 9 7 nd ()(7), we hve perfect squre trinomil: 9 ( 7) c) Becuse (), 9, nd the middle term is equl to ()(), the trinomil 9 is perfect squre trinomil: 9 ( ) Fctoring Difference or Sum of Two Cubes In Emple 6 of Section 5.5 we divided b by b to get the quotient b b nd no reminder. So b is fctor of b, difference of two cubes. If you divide b by b, you will get the quotient b b nd no reminder. Try it. So b is fctor of b, sum of two cubes. These results give us two more fctoring rules. Fctoring Difference or Sum of Two Cubes b ( b)( b b) b ( b)( b b) 7 Fctoring difference or sum of two cubes Fctor ech polynomil. b) y ) 8 c) 8z 7 ) Becuse 8, we cn use the formul for fctoring the difference of two cubes. In the formul b ( b)( b b), let nd b : 8 ( )( ) Recognize sum of two cubes. b) y y ( y )(y y ) Let y nd b in the formul for the sum of two cubes.

45 5.6 c) 8z 7 (z) (z )(z 6z 9) Fctoring Polynomils (5-5) 0 Recognize difference of two cubes. Let z nd b in the formul for difference of two cubes. Fctoring Polynomil Completely Polynomils tht cnnot be fctored re clled prime polynomils. Becuse binomils such s 5, 6, nd cnnot be fctored, they re prime polynomils. A polynomil is fctored completely when it is written s product of prime polynomils. To fctor completely, lwys fctor out the GCF (or its opposite) first. Then continue to fctor until ll of the fctors re prime. 8 Fctoring completely Fctor ech polynomil completely. b) 0 75 ) 5 0 c) b 6b Gretest common fctor ) 5 0 5( ) 5( )( ) Difference of two squres b) 0 75 ( 0 5) Gretest common fctor ( 5) Perfect squre trinomil c) b 6b b(b 8) Fctor out b to mke the net step esier. b(b )(b b ) Difference of two cubes Fctoring by Substitution So fr, the polynomils tht we hve fctored, without common fctors, hve ll been of degree or. Some polynomils of higher degree cn be fctored by substituting single vrible for vrible with higher power. After fctoring, we replce the single vrible by the higher-power vrible. This method is clled substitution. helpful 9 Fctoring by substitution Fctor ech polynomil. ) 9 ) We recognize 9 s difference of two squres in which ( ) nd 9. If we let w, then w. So we cn replce by w nd fctor: Replce by w. 9 w 9 (w )(w ) Difference of two squres ( )( ) Replce w by. b) We recognize y8 y 9 s perfect squre trinomil in which y8 ( y ) nd 9 7. We let w y nd w y8: y8 y 9 w w 9 Replce y by w nd y8 by w. (w 7) Perfect squre trinomil ( y 7) Replce w by y. hint It is not ctully necessry to perform the substitution step. If you cn recognize tht 9 ( )( ) then skip the substitution. b) y8 y 9

46 0 (5-6) Chpter 5 Eponents nd Polynomils CAUTION The polynomils tht we fctor by substitution must contin just the right powers of the vrible. We cn fctor y8 y 9 becuse ( y) y8, but we cnnot fctor y7 y 9 by substitution. In the net emple we use substitution to fctor polynomils tht hve vribles s eponents. 0 Polynomils with vrible eponents Fctor completely. The vribles used in the eponents represent positive integers. b) z n 6z n 9z ) m y ) Notice tht m ( m). So if we let w m, then w m: Substitution m y w y (w y)(w y) Difference of two squres m m ( y)( y) Replce w by m. b) First fctor out the common fctor z: z n 6z n 9z z(zn 6z n 9) z ( 6 9) Let z n. z ( ) Perfect squre trinomil n z (z ) Replce by z n. WARM-UPS True or flse? Eplin your nswer For the polynomil y 6y we cn fctor out either y or y. True The gretest common fctor for the polynomil 8 5b is. True ( ) for ny vlue of. True 6 ( )( ) for ny vlue of. True The polynomil 6 6 is perfect squre trinomil. Flse The polynomil y 6 is perfect squre trinomil. Flse 9 9 ( 7) for ny vlue of. Flse The polynomil is fctor of. True 7 ( )( 6 9) for ny vlue of. Flse 8 ( ) for ny vlue of. Flse EXERCISES Reding nd Writing After reding this section, write out the nswers to these questions. Use complete sentences.. Wht is prime number? A prime number is nturl number greter thn tht hs no fctors other thn itself nd.. When is nturl number fctored completely? A nturl number is fctored completely when it is epressed s product of prime numbers.. Wht is the gretest common fctor for the terms of polynomil? The gretest common fctor for the terms of polynomil is monomil tht includes every number or vrible tht is fctor of ll of the terms of the polynomil.. Wht re the two wys to fctor out the gretest common fctor? The gretest common fctor cn be fctored out with positive coefficient or negtive coefficient.

47 Wht is qudrtic polynomil? 7. Wht is prime polynomil? 8. When is polynomil fctored completely? Find the gretest common fctor for ech group of terms. See Emple. 9. 8, 6 0., 8. 9w, wy, 5y. 70, 8,. y, y, 66y. 60b5, 09b, 0b6 Fctor out the gretest common fctor in ech epression. See Emples nd ( y 7. 8w 6wy 8. wz 8w b6 b 60 5b. 8y6z 0 6y 9z. ( 6) ( 6)b. (y ) ( y )b 5. (y ) y ( y )z 6. (w ) w (w ) Fctor out the gretest common fctor, then fctor out the opposite of the gretest common fctor. See Emple. 7. y w w. w 6w Fctor ech polynomil. See Emple y 7. y b (5-7) b. w z. y 9c Fctor ech polynomil. See Emple y 0y 5 5. m m 6. 9t 0t 5 7. w wt t 8. r 0rt 5t Fctor. See Emple w 5. w m 8 Fctor ech polynomil completely. See Emple m 5 m ( ) ( )7 6. ( ) ( ) y y 66. y y m mn y y 5 7. ( ) ( ) 7. ( ) ( ) 7. 9 w ( w 7. bn b n b y 6 y 79. y 8y 7y 80. m n 8mn 8n 8. 7 b Wht is liner polynomil? Fctoring Polynomils 0

48 0 (5-8) Chpter 5 Eponents nd Polynomils Fctor ech polynomil completely. See Emple 9. c) Use the ccompnying grph to estimte the height of cge for which the volume is 0,000 cubic inches y8 85. z 6z Volume (in thousnds of cubic inches) y Fctor ech polynomil completely. The vribles used s eponents represent positive integers. See Emple n Height of cge (in inches) 50 FIGURE FOR EXERCISE bn r 6 r 9 6. Pyrmid power. A powerful crystl pyrmid hs squre bse nd volume of y y y cubic centimeters. If its height is y centimeters, then wht polynomil represents the length of side of the squre bse? The volume of pyrmid with squre bse of re h nd height h is given by V. 98. u6n un 99. m yn 0. m b 0. r m 8t 0. k w 0k w 5k 0. t t 05. uv6k uvk uv k 06. umv n u mv n u mv n y Replce k in ech trinomil by number tht mkes the trinomil perfect squre trinomil k 08. y 8y k 09. k u kuv 9v. km m 9. kz 0z 6. 8y 80y k k Solve ech problem. FIGURE FOR EXERCISE 6 GET TING MORE INVOLVED 5. Volume of bird cge. A compny mkes rectngulr shped bird cges with height b inches nd squre bottoms. The volume of these cges is given by the function V b 6b 9b. ) Wht is the length of side of the squre bottom? b) Use the function to find the volume of cge with height of 8 inches. 7. Coopertive lerning. List the perfect squre trinomils corresponding to ( ), ( ), ( ),..., ( ). Use your list to quiz clssmte. Red perfect squre trinomil t rndom from your list nd sk your clssmte to write its fctored form. Repet until both of you hve mstered these perfect squre trinomils.

49 5.7 section (5-9) 05 F A C T O R I N G b c 5.7 In this Fctoring b c In Section 5.5 you lerned to fctor certin specil polynomils. In this section you will lern to fctor generl qudrtic polynomils. We first fctor b c with, nd then we consider the cse. Fctoring Trinomils with Leding Coefficient Fctoring Trinomils with Leding Coefficient Fctoring Trinomils with Leding Coefficient Not Let s look closely t n emple of finding the product of two binomils using the distributive property: Tril nd Error Higher Degrees nd Vrible Eponents ( )( ) ( ) ( ) 7 Distributive property Distributive property To fctor 7, we need to reverse these steps. First observe tht the coefficient 7 is the sum of two numbers tht hve product of. The only numbers tht hve product of nd sum of 7 re nd. So write 7 s : 7 Now fctor the common fctor out of the first two terms nd the common fctor out of the lst two terms. This method is clled fctoring by grouping. Fctor out Fctor out 7 ( ) ( ) ( )( ) Fctoring trinomil by grouping Fctor ech trinomil by grouping. ) 9 8 Rewrite 7 s. Fctor out common fctors. Fctor out the common fctor. b) ) We need to find two integers with product of 8 nd sum of 9. For product of 8 we could use nd 8, nd 9, or nd 6. Only nd 6 hve sum of 9. So we replce 9 with 6 nd fctor by grouping: Replce 9 by 6. ( ) ( )6 Fctor out common fctors. ( )( 6) Check by using FOIL. b) We need to find two integers with product of nd sum of. For product of we hve nd, nd, nd 8, or nd 6. To get product of nd sum of, we must use nd 6: 6 Replce with 6. ( 6) ( 6) Fctor out common fctors. ( 6)( ) Check by using FOIL. The method shown in Emple cn be shortened gretly. Once we discover tht nd 6 hve product of 8 nd sum of 9, we cn simply write 9 8 ( )( 6).

50 06 (5-50) Chpter 5 Eponents nd Polynomils Once we discover tht nd 6 hve product of nd sum of, we cn simply write ( 6)( ). In the net emple we use this shortcut. study Fctoring b c with Fctor ech qudrtic polynomil. b) 0 ) c) 5 6 ) Two integers with product of nd sum of re nd : ( )( ) Check by using FOIL. b) Two integers with product of 0 nd sum of re 5 nd : 0 ( 5)( ) Check by using FOIL. c) Two integers with product of 6 nd sum of 5 re nd : 5 6 ( )( ) tip Set short-term gols nd rewrd yourself for ccomplishing them. When you hve solved 0 problems, tke short brek nd listen to your fvorite music. Check by using FOIL. Fctoring Trinomils with Leding Coefficient Not If the leding coefficient of qudrtic trinomil is not, we cn gin use grouping to fctor the trinomil. However, the procedure is slightly different. Consider the trinomil, for which, b, nd c. First find c, the product of the leding coefficient nd the constnt term. In this cse c. Now find two integers with product of nd sum of. The pirs of integers with product of re nd, nd, nd 8, nd nd 6. Only nd 8 hve product of nd sum of. Now replce by 8 nd fctor by grouping: 8 ( ) ( ) ( )( ) This strtegy for fctoring qudrtic trinomil, known s the c method, is summrized in the following bo. The c method works lso when. Strtegy for Fctoring b c by the c Method To fctor the trinomil b c. find two integers tht hve product equl to c nd sum equl to b,. replce b by two terms using the two new integers s coefficients,. then fctor the resulting four-term polynomil by grouping.

51 5.7 Fctoring b c with Fctor ech trinomil. ) 9 Fctoring b c (5-5) 07 b) 5 ) Becuse 8, we need two numbers with product of 8 nd sum of 9. The numbers re nd 8. Replce 9 by 8 nd fctor by grouping: 9 8 ( ) ( ) ( )( ) Check by FOIL. Note tht if you strt with 8, nd fctor by grouping, you get the sme result. b) Becuse ( ), we need two numbers with product of nd sum of 5. The pirs of numbers with product of re nd, nd, nd 8, nd nd 6. To get product of, one of the numbers must be negtive nd the other positive. To get sum of positive 5, we need nd 8: 5 8 ( ) ( ) ( )( ) Check by FOIL. Tril nd Error helpful After we hve gined some eperience t fctoring by grouping, we cn often find the fctors without going through the steps of grouping. Consider the polynomil hint The c method hs more written work nd less guesswork thn tril nd error. However, mny students enjoy the chllenge of trying to write only the nswer without ny other written work The fctors of cn only be nd. The fctors of 6 could be nd or nd 6. We cn list ll of the possibilities tht give the correct first nd lst terms without putting in the signs: ( )( ) ( )( ) ( 6)( ) ( )( 6) Before ctully trying these out, we mke n importnt observtion. If ( ) or ( 6) were one of the fctors, then there would be common fctor in the originl trinomil, but there is not. If the originl trinomil hs no common fctor, there cn be no common fctor in either of its liner fctors. Since 6 is positive nd the middle term is 7, both of the missing signs must be negtive. So the only possibilities re ( )( 6) nd ( )( ). The middle term of the first product is, nd the middle term of the second product is 7. So we hve found the fctors: 7 6 ( )( ) Even though there my be mny possibilities in some fctoring problems, often we find the correct fctors without writing down every possibility. We cn use bit of guesswork in fctoring trinomils. Try whichever possibility you think might work. Check it by multiplying. If it is not right, then try gin. Tht is why this method is clled tril nd error.

52 08 (5-5) Chpter 5 Eponents nd Polynomils Tril nd error Fctor ech qudrtic trinomil using tril nd error. b) 6 ) 5 ) Becuse fctors only s nd fctors only s, there re only two possible wys to fctor this trinomil to get the correct first nd lst terms: ( )( ) nd ( )( ) Becuse the lst term of the trinomil is negtive, one of the missing signs must be, nd the other must be. Now we try the vrious possibilities until we get the correct middle term: ( )( ) 5 ( )( ) ( )( ) 5 Since the lst product hs the correct middle term, the trinomil is fctored s 5 ( )( ). b) There re four possible wys to fctor 6: ( )( 6) ( )( ) ( 6)( ) ( )( ) Becuse the lst term is positive nd the middle term is negtive, both signs must be negtive. Now try possible fctors until we get the correct middle term: ( )( 6) 9 6 ( )( ) 6 The trinomil is fctored correctly s 6 ( )( ). Higher Degrees nd Vrible Eponents It is not necessry lwys to use substitution to fctor polynomils with higher degrees or vrible eponents s we did in Section 5.6. In the net emple we use tril nd error to fctor two polynomils of higher degree nd one with vrible eponents. Remember tht if there is common fctor to ll terms, fctor it out first. 5 Higher-degree nd vrible eponent trinomils Fctor ech polynomil completely. Vribles used s eponents represent positive integers. b) 8y7 y 5y c) um 5um ) 8 5 ) To fctor by tril nd error, notice tht 8. Now 5 is 5 or 5. Using nd 5 will not give the required for the coefficient of the middle term. So choose nd 5 to get the in the middle term: 8 5 ( 5)( ) b) 8y7 y 5y y(6y6 7y 5) Fctor out the common fctor y first. y(y )(y 5) Fctor the trinomil by tril nd error. m m c) Notice tht u u um nd. Using tril nd error, we get um 5um (um )(um ).

53 5.7 WARM-UPS (5-5) 09 True or flse? Answer true if the polynomil is fctored correctly nd flse otherwise Fctoring b c 9 8 ( )( 6) y y 5 (y 5)(y 7) ( )( ) 5 6 ( )( ) ( 6)( ) 5 6 ( )( 9) 5 ( 5)( ) ( )( ) ( )( ) 8 ( )( ) EXERCISES Fctor ech polynomil using the c method. See Emple. 7. 6w 5w 8. 6 Reding nd Writing After reding this section, write out the nswers to these questions. Use complete sentences.. How do we fctor trinomils tht hve leding coefficient of?. How do we fctor trinomils in which the leding coefficient is not? y 7y m m 5. y y b b. Wht is tril-nd-error fctoring? Fctor ech polynomil using tril nd error. See Emple Wht should you lwys first look for when fctoring polynomil? Fctor ech polynomil. See Emples nd y 5y t t. b 6b 5. y 7y. 6w 5w m m t 9t 9. 6y 7y u u 6 9. y 5y y y Fctor ech polynomil completely. See Emple 5. The vribles used in eponents represent positive integers w 9w b6 b8

54 0 (5-5) Chpter 5 Eponents nd Polynomils b7 b b 75. 9m 5n 76. m n mn n y b y b y 9y 0 9. y b 50. w m 79. w 8w z yz y z m 0 5m w y y w 9w y b y b w w y z 6 5z y 6y 58. u 6 v 6 5u v u m w 0w m m 9. 6w z 6z Fctor ech polynomil completely y y y m 5 0m 00m y y 6 7. y y 6 7. m m 5.8 section Prime Polynomils Fctoring Polynomils Completely Fctoring Polynomils with Four Terms Summry 95. Discussion. Which of the following is not perfect squre trinomil? Eplin. ) 6 6 b 9b8 b) c) 900y 60y d) 6 6z 7 9z 96. Discussion. Which of the following is not difference of two squres? Eplin. ) 6 8y 5c b) 9 b 90 c) t d) 96 FACTORING STRATEGY In previous sections we estblished the generl ide of fctoring nd mny specil cses. In this section we will see tht polynomil cn hve s mny fctors s its degree, nd we will fctor higher-degree polynomils completely. We will lso see generl strtegy for fctoring polynomils. In this GET TING MORE INVOLVED Prime Polynomils A polynomil tht cnnot be fctored is prime polynomil. Binomils with no common fctors, such s nd, re prime polynomils. To determine whether polynomil such s is prime polynomil, we must try ll possibilities for fctoring it. If could be fctored s product of two binomils, the only possibilities tht would give first term of nd lst term of re ( )( ) nd ( )( ). However, ( )( ) nd ( )( ).

55 5.8 Fctoring Strtegy (5-55) Both products hve n -term. Of course, ( )( ) hs no -term, but ( )( ). Becuse none of these possibilities results in, the polynomil is prime polynomil. Note tht is sum of two squres. A sum of two squres of the form b is lwys prime polynomil. Prime polynomils Determine whether the polynomil is prime polynomil. To fctor, we must find two integers with product of nd sum of. The only pirs of positive integers with product of re nd, nd nd. Becuse the product is positive, both numbers must be negtive or both positive. Under these conditions it is impossible to get sum of positive. The polynomil is prime. Fctoring Polynomils Completely So fr, typicl polynomil hs been product of two fctors, with possibly common fctor removed first. However, it is possible tht the fctors cn still be fctored gin. A polynomil in single vrible my hve s mny fctors s its degree. We hve fctored polynomil completely when ll of the fctors re prime polynomils. Fctoring higher-degree polynomils completely Fctor completely. Two numbers with product of nd sum of re nd : ( )( ) ( )( )( ) Difference of two squres Since,, nd re prime, the polynomil is fctored completely. In the net emple we fctor sith-degree polynomil. study Fctoring completely Fctor 6 completely. To fctor 6, we must first fctor out the common fctor nd then recognize tht 6 is perfect squre: 6 ( ) : Fctor out the common fctor. 6 ( 6 ) (( ) ) Write 6 s perfect squre. ( )( ) Difference of two squres ( )( )( )( ) Difference of two cubes nd tip Find clen, comfortble,welllit plce to study. But don t get too comfortble. Sitting t desk is preferble to lying in bed. Where you study cn influence your concentrtion nd your study hbits. sum of two cubes Since nd re prime, the polynomil is fctored completely.

56 (5-56) Chpter 5 Eponents nd Polynomils In Emple we recognized 6 s difference of two squres. However, is lso difference of two cubes, nd we cn fctor it using the rule for the difference of two cubes: 6 6 ( ) ( )( ) Now we cn fctor, but it is difficult to see how to fctor. (It is not prime.) Although 6 cn be thought of s perfect squre or perfect cube, in this cse thinking of it s perfect squre is better. In the net emple we use substitution to simplify the polynomil before fctoring. This fourth-degree polynomil hs four fctors. Using substitution to simplify Fctor (w ) (w ) completely. Let w to simplify the polynomil: Replce w by. (w ) (w ) ( 8)( ) (w 8)(w ) Replce by w. (w 9)(w ) (w )(w )(w )(w ) Fctoring Polynomils with Four Terms In Section 5.6 we rewrote trinomil s polynomil with four terms nd then used fctoring by grouping. Fctoring by grouping cn lso be used on other types of polynomils with four terms. 5 Polynomils with four terms Use grouping to fctor ech polynomil completely. b) 7 9 ) c) bw b w ) Note tht the first two terms of hve common fctor of, nd the lst two terms hve common fctor of. ( ) ( ) Fctor by grouping. ( )( ) Fctor out. Since is sum of two squres, it is prime nd the polynomil is fctored completely. b) We cn fctor out of the first two terms of 7 9 nd 9 or 9 from the lst two terms. We choose 9 to get the fctor in ech cse. 7 9 ( ) 9( ) Fctor by grouping. ( 9)( ) Fctor out. ( )( )( ) Difference of two squres This third-degree polynomil hs three fctors.

57 5.8 Fctoring Strtegy (5-57) c) First rerrnge the terms so tht the first two nd the lst two hve common fctors: Rerrnge the terms. bw b w b w bw ( b) w( b) Common fctors Fctor out b. ( w)( b) Summry A strtegy for fctoring polynomils is given in the following bo. Strtegy for Fctoring Polynomils. If there re ny common fctors, fctor them out first.. When fctoring binomil, look for the specil cses: difference of two squres, difference of two cubes, nd sum of two cubes. Remember tht sum of two squres b is prime.. When fctoring trinomil, check to see whether it is perfect squre trinomil.. When fctoring trinomil tht is not perfect squre, use grouping or tril nd error. 5. When fctoring polynomil of high degree, use substitution to get polynomil of degree or, or use tril nd error. 6. If the polynomil hs four terms, try fctoring by grouping. helpful 6 Using the fctoring strtegy Fctor ech polynomil completely. ) w w 8w c) 6 b 80b 00b ) The gretest common fctor (GCF) for the three terms is w: Fctor out w. w w 8w w(w w 6) w(w )(w ) Fctor completely. b) The GCF in 0 60 is 0: ( 6) Becuse 6 is prime, the polynomil is fctored completely. c) The GCF in 6 b 80b 00b is b: 6 b 80b 00b b( 0 5) b( 5) d) The polynomil hs four terms, nd we cn fctor it by grouping: w mw z mz w( m) z( m) (w z)( m) hint When fctoring integers, we write. However, when fctoring polynomils we usully do not fctor ny of the integers tht pper. So we sy tht b( 5) is fctored completely. b) 0 60 d) w mw z mz

58 (5-58) Chpter 5 Eponents nd Polynomils WARM-UPS True or flse? Eplin your nswer ( ) for ny vlue of. Flse The polynomil 9 is perfect squre trinomil. True The sum of two squres b is prime. True The polynomil 6 is fctored completely s ( )( ). Flse y 7 (y )(y y 9) for ny vlue of y. Flse The polynomil y 6 is difference of two squres. True The polynomil is fctored completely s ( )( ). Flse The polynomil is prime polynomil. True The polynomil 6 is the difference of two cubes. True The polynomil cn be fctored by grouping EXERCISES Reding nd Writing After reding this section, write out the nswers to these questions. Use complete sentences.. ( 5). ( ). Wht should you do first when fctoring polynomil? Alwys fctor out the gretest common fctor first.. y 6 8. If you re fctoring binomil, then wht should you look for? When fctoring trinomil wht should you look for?. 6 6y ( 6) ( 8. y (y ) ( 9. (m ) (m ). Wht should you look for when fctoring four-term polynomil?. (y ) (y ) 0. (w ) 5(w ) 5. y 00 Prime w y 6 Not prime 9. Not prime Prime. (m 8) (m 8) (m )(m )(m )(m ) Use grouping to fctor ech polynomil completely. See Emple 5. Not prime 0. Prime. Prime. Not prime 5. y b by (. Prime. Prime z k kz ( Prime Prime ( Fctor ech polynomil completely. See Emples w bw b 8. 9y y 0. w wy y c bc 5 5b (w. (y ) ( y ) (y )(y )(y )(y ) Determine whether ech polynomil is prime polynomil. See Emple. (m 0. (w ) (w ) 5 ( ( (c

59 5.8. y 5y 8y 0 Fctoring Strtegy 8. w 5. dy d w wy 8. (w 5) 9 6. y by b 8. ( 6) 7. y y 85. w w 9 8. d b c c b d 86. 9n 5n n 9. y b by y b mw bm w Use the fctoring strtegy to fctor ech polynomil completely. See Emple m n n m 6m Fctor completely. Assume vribles used s eponents represent positive integers m ( y) 99. w b 6n n 9 6. b b 0. t n 6 6. m n n n ( 5) 0. m m 6 m (5-59) 67. m n mn n 05. n n nb b 68. b 6b 9b 06. mz 5z m m wn n wm 70. w 5b bw 5 GET TING MORE INVOLVED 7. w w 07. Coopertive lerning. Write down 0 trinomils of the form b c t rndom using integers for, b, nd c. Wht percent of your 0 trinomils re prime? Would you sy tht prime trinomils re the eception or the rule? Compre your results with those of your clssmtes. 7. w 8w 6 7. t t 7. m 5m y y 0y 08. Writing. The polynomil 77. (y 5) (y 5) (t ) 7(t ) 0 is product of five fctors of the form n, where n is nturl number smller thn. Fctor this polynomil completely nd eplin your procedure. 79. w

60 6 (5-60) Chpter 5 Eponents nd Polynomils 5.9 The techniques of fctoring cn be used to solve equtions involving polynomils tht cnnot be solved by the other methods tht you hve lerned. After you lern to solve equtions by fctoring, we will use this technique to solve some new pplied problems in this section nd in Chpters 6 nd 8. In this section The Zero Fctor Property Applictions SOLVING EQUATIONS BY FACTORING The Zero Fctor Property The eqution b 0 indictes tht the product of two unknown numbers is 0. But the product of two rel numbers is zero only when one or the other of the numbers is 0. So even though we do not know ectly the vlues of nd b from b 0, we do know tht 0 or b 0. This ide is clled the zero fctor property. helpful hint Zero Fctor Property Note tht the zero fctor property is our second emple of getting n equivlent eqution without doing the sme thing to ech side. Wht ws the first? 0 or b 0. The net emple shows how to use the zero fctor property to solve n eqution in one vrible. Using the zero fctor property Solve 0. We fctor the left-hnd side of the eqution to get product of two fctors tht re equl to 0. Then we write n equivlent eqution using the zero fctor property. 0 ( )( ) 0 Fctor the left-hnd side. 0 or 0 Zero fctor property or Solve ech prt of the compound eqution. Check tht both nd stisfy 0. If, we get () () 6 0. If, we get () 9 0. So the solution set is,. The zero fctor property is used only in solving polynomil equtions tht hve zero on one side nd polynomil tht cn be fctored on the other side. The polynomils tht we fctored most often were the qudrtic polynomils. The equtions tht we will solve most often using the zero fctor property will be qudrtic equtions. Qudrtic Eqution If, b, nd c re rel numbers, with 0, then the eqution b c 0 is clled qudrtic eqution. The eqution b 0 is equivlent to the compound eqution

61 5.9 M A T H A T Solving Equtions by Fctoring (5-6) 7 W O R K Sems Mercdo, professionl bodyborder nd 988 Ntionl Chmpion, chrges the wves off Hwii, Thiti, Indonesi, Meico, nd Cliforni. In choosing bord for competition nd for the mneuvers he wnts to perform, Mercdo fctors in his height nd weight s well s the size, power, nd temperture of the wves he will be riding. In colder wter softer, BODYBOARD more fleible bord is used; in wrmer wter stiffer bord is DESIGNER chosen. When wves crsh on shore, the ride usully lsts to 5 seconds, nd shorter bord with nrrow til is chosen for greter control. When wves brek long snd br or reef, the ride cn sometimes lst s long s minutes, nd strighter bord with more surfce re is chosen so tht the bord will move fster nd llow the rider to pull off more mneuvers. Bsic mneuvers include bottom turns, erils, forwrd nd reverse 60 s, nd el rollos. As one of the top 0 bodyborders in the world, Mercdo helps to design the bords he uses. Performnce levels re gretly incresed with fine-tuned equipment nd techniques, he sys. In Eercise 6 of Section 5.9 you will find the dimensions of given bodybord. In Chpter 8 we will study qudrtic equtions further nd solve qudrtic equtions tht cnnot be solved by fctoring. Keep the following strtegy in mind when solving equtions by fctoring. study tip Strtegy for Solving Equtions by Fctoring We re ll cretures of hbit. When you find plce in which you study successfully, stick with it. Using the sme plce for studying will help you to concentrte nd to ssocite the plce with good studying.. Write the eqution with 0 on the right-hnd side.. Fctor the left-hnd side.. Use the zero fctor property to get simpler equtions. (Set ech fctor equl to 0.). Solve the simpler equtions. 5. Check the nswers in the originl eqution. Solving qudrtic eqution by fctoring Solve ech eqution. b) 6 9 ) 0 5 ) Use the steps in the strtegy for solving equtions by fctoring: 0 5 Originl eqution Rewrite with zero on the right-hnd side. 5( ) 0 Fctor the left-hnd side. 5 0 or 0 Zero fctor property 0 or Solve for. The solution set is 0,. Check ech solution in the originl eqution.

62 8 (5-6) Chpter 5 Eponents nd Polynomils b) First rewrite the eqution with 0 on the right-hnd side nd the left-hnd side in order of descending eponents: 6 9 Originl eqution Add 9 to ech side. Divide ech side by. 0 Fctor. ( )( ) 0 Zero fctor property 0 or 0 or Solve for. The solution set is,. Check ech solution in the originl eqution. CAUTION If we divide ech side of 0 5 by 5, we get, or. We do not get 0. By dividing by 5 we hve lost one of the fctors nd one of the solutions. In the net emple there re more thn two fctors, but we cn still write n equivlent eqution by setting ech fctor equl to 0. Solving cubic eqution by fctoring Solve 8 0. clcultor First notice tht the first two terms hve the common fctor nd the lst two terms hve the common fctor. close-up To check, use Y= to enter y 8. Then use the vribles feture (VARS) to find y(), y( ), nd y(). 0 or or ( ) ( ) 0 ( )( ) 0 ( )( )( ) 0 0 or 0 or Fctor by grouping. Fctor out. Fctor completely. Set ech fctor equl to 0. The solution set is,,. Check ech solution in the originl eqution. The eqution in the net emple involves bsolute vlue. Solving n bsolute vlue eqution by fctoring Solve 6 8. First write n equivlent compound eqution without bsolute vlue: or or ( 6)( ) 0 or ( )( ) or 0 or 0 or 0 6 or or or The solution set is,,, 6. Check ech solution.

63 5.9 Solving Equtions by Fctoring (5-6) 9 Applictions Mny pplied problems cn be solved by using equtions such s those we hve been solving. helpful 5 Are of room Ronld s living room is feet longer thn it is wide, nd its re is 68 squre feet. Wht re the dimensions of the room? Let be the width nd be the length. See Figure 5.. Becuse the re of rectngle is the length times the width, we cn write the eqution ( ) 68. hint To prove the Pythgoren theorem, drw two squres with sides of length b, nd prtition them s shown. b c b b b c b b FIGURE 5. We solve the eqution by fctoring: c c 68 0 ( )( ) 0 0 or 0 or Becuse the width of room is positive number, we disregrd the solution. We use nd get width of feet nd length of feet. Check this nswer by multiplying nd to get 68. b c c b c b Erse the four tringles in ech picture. Since we strted with equl res, we must hve equl res fter ersing the tringles: Applictions involving qudrtic equtions often require theorem clled the Pythgoren theorem. This theorem sttes tht in ny right tringle the sum of the squres of the lengths of the legs is equl to the length of the hypotenuse squred. The Pythgoren Theorem The tringle shown is right tringle if nd only if b c. Hypotenuse c b Legs We use the Pythgoren theorem in the net emple. b c

64 0 (5-6) Chpter 5 Eponents nd Polynomils 6 Using the Pythgoren theorem Shirley used meters of fencing to enclose rectngulr region. To be sure tht the region ws rectngle, she mesured the digonls nd found tht they were 5 meters ech. (If the opposite sides of qudrilterl re equl nd the digonls re equl, then the qudrilterl is rectngle.) Wht re the length nd width of the rectngle? The perimeter of rectngle is twice the length plus twice the width, P L W. Becuse the perimeter is meters, the sum of one length nd one width is 7 meters. If we let represent the width, then 7 is the length. We use the Pythgoren theorem to get reltionship mong the length, width, nd digonl. See Figure FIGURE 5. (7 ) ( )( ) 0 0 or 0 or 7 or 7 Pythgoren theorem Simplify. Simplify. Divide ech side by. Fctor the left-hnd side. Zero fctor property Solving the eqution gives two possible rectngles: by rectngle or by rectngle. However, those re identicl rectngles. The rectngle is meters by WARM-UPS True or flse? Eplin your nswer.. The eqution ( )( ) is equivlent to or. Flse. Equtions solved by fctoring my hve two solutions. True. The eqution c d 0 is equivlent to c 0 or d 0. True. The eqution 5 is equivlent to the compound eqution 5 or 5. Flse 5. The solution set to the eqution ( )( ) 0 is,. True 6. The Pythgoren theorem sttes tht the sum of the squres of ny two sides of ny tringle is equl to the squre of the third side. Flse 7. If the perimeter of rectngulr room is 8 feet, then the sum of the length nd width is 9 feet. True 8. Two numbers tht hve sum of 8 cn be represented by nd 8. True 9. The solution set to the eqution ( )( ) 0 is,. Flse 0. The solution set to the eqution ( )( 5) 0 is,, 5. Flse

65 Solving Equtions by Fctoring (5-65) EXERCISES , Solve ech eqution. Reding nd Writing After reding this section, write out the nswers to these questions. Use complete sentences.. Wht is the zero fctor property?. Wht is qudrtic eqution?. Where is the hypotenuse in right tringle? 5 5, Where re the legs in right tringle? Wht is the Pythgoren theorem? 6. Where is the digonl of rectngle? Solve ech eqution. See Emples. 7. ( 5)( ) 0 8. ( 6)( 5) , , 5 6 ( )( ) ( )( ) 0 y 9y 0y 0 m m m 0 5 5, 0, 5 5 5, 0, 5 7. ( )( 9) 0 9. ( 5)( ) 0 8. ( )( )( 9) 0 0. (k 8)(k ) w 5w 0 t 6t 7 0 m 7m 0 0, 7 h 5h 0 0, 5 0, 5 p p 6, z z 0, m m 0, 0, 6 0 w w 5w ,, 5 6 0,, n n n 0,, w w 5w 5 0 5,, Solve ech eqution for y. Assume nd b re positive numbers. 5. y by 0 0, b 5. y y by b 0 5. y b y 6y y b 0 b, b 57. y y y 58. y by b 0 Solve ech problem. See Emples 5 nd Color print. The length of new super size color print is inches more thn the width. If the re is squre inches, wht re the length nd width? Width inches, length 6 inches Solve ech eqution. See Emple. 7. 5,,, 55. y by b 0

66 (5-66) Chpter 5 Eponents nd Polynomils 60. Tennis court dimensions. In singles competition, ech plyer plys on rectngulr re of 7 squre yrds. Given tht the length of tht re is yrds greter thn its width, find the length nd width. Width 9 yrds, length yrds 6. Missing numbers. The sum of two numbers is nd their product is 6. Find the numbers. nd 9 6. More missing numbers. The sum of two numbers is 6.5, nd their product is 9. Find the numbers. nd.5 6. Bodybording. The Sems Chnnel pro bodybord shown in the figure hs length tht is inches greter thn its width. Any rider weighing up to 00 pounds cn use it becuse its surfce re is 96 squre inches. Find the length nd width. Length inches, width inches c) Use the ccompnying grph to estimte the mimum height reched by the rrow. 6 feet d) At wht time does the rrow rech its mimum height? seconds Height (feet) Time (seconds) 5 FIGURE FOR EXERCISE Time until impct. If n object is dropped from height of s0 feet, then its ltitude fter t seconds is given by the formul S 6t s0. If pck of emergency supplies is dropped from n irplne t height of 600 feet, then how long does it tke for it to rech the ground? 0 seconds 67. Yolnd s closet. The length of Yolnd s closet is feet longer thn twice its width. If the digonl mesures feet, then wht re the length nd width? Width 5 feet, length feet in. in. FIGURE FOR EXERCISE 6 6. New dimensions in grdening. Mry Gold hs rectngulr flower bed tht mesures feet by 6 feet. If she wnts to increse the length nd width by the sme mount to hve flower bed of 8 squre feet, then wht will be the new dimensions? 6 feet by 8 feet ft ft ft ft FIGURE FOR EXERCISE Ski jump. The bse of ski rmp forms right tringle. One leg of the tringle is meters longer thn the other. If the hypotenuse is 0 meters, then wht re the lengths of the legs? 6 feet nd 8 feet ft 6 ft ft 0 m FIGURE FOR EXERCISE Shooting rrows. An rcher shoots n rrow stright upwrd t 6 feet per second. The height of the rrow h(t) (in feet) t time t seconds is given by the function h(t) 6t 6t. ) Use the ccompnying grph to estimte the mount of time tht the rrow is in the ir. seconds b) Algebriclly find the mount of time tht the rrow is in the ir. seconds m +m FIGURE FOR EXERCISE Trimming gte. A totl of feet of lumber is used round the perimeter of the gte shown in the figure on the net pge. If the digonl brce is feet long, then wht re the length nd width of the gte? Width 5 feet, length feet

67 Chpter 5 Collbortive Activities (5-67) 7. Arrnging the rows. Mr. Converse hs students in his lgebr clss with n equl number in ech row. If he rrnges the desks so tht he hs one fewer rows, he will hve two more students in ech row. How mny rows did he hve originlly? 8 f t GET TING MORE INVOLVED 75. Writing. If you divide ech side of by, you get. If you subtrct from ech side of, you get 0, which hs two solutions. Which method is correct? Eplin. FIGURE FOR EXERCISE Perimeter of rectngle. The perimeter of rectngle is 8 inches, nd the digonl mesures 0 inches. Wht re the length nd width of the rectngle? Length 8 inches, width 6 inches 7. Consecutive integers. The sum of the squres of two consecutive integers is 5. Find the integers. nd, or nd 7. Pete s grden. Ech row in Pete s grden is feet wide. If the rows run north nd south, he cn hve two more rows thn if they run est nd west. If the re of Pete s grden is 5 squre feet, then wht re the length nd width? Length 5 feet, width 9 feet 7. House plns. In the plns for their drem house the Bileys hve mster bedroom tht is 0 squre feet in re. If they increse the width by feet, they must decrese the length by feet to keep the originl re. Wht re the originl dimensions of the bedroom? Length 0 feet, width feet 76. Coopertive lerning. Work with group to emine the following solution to : ( ) or or Is this method correct? Eplin. 77. Coopertive lerning. Work with group to emine the following steps in the solution to ( ) 0 5( )( ) 0 0 or 0 or Wht hppened to the 5? Eplin. COLLABORATIVE ACTIVITIES Mgic Tricks Grouping: Two students per group Topic: Prctice with eponent rules, multiplying polynomils Jim nd Sdr re tlking one dy fter clss. Sdr: Jim, I hve trick for you. Think of number between nd 0. I will sk you to do some things to this number. Then t the end tell me your result, nd I will tell you your number. Jim: Oh, yeh you probbly rig it so the result is my number. Sdr: Come on Jim, give it try nd see. Jim: Oky, oky, I thought of number. Sdr: Good, now write it down, nd don t let me see your pper. Now dd. Got tht? Now multiply everything by. Jim: Hey, I didn t know you were going to mke me think! This is lgebr! Sdr: I know, now just do it. Oky, now squre the polynomil. Got tht? Now subtrct. Jim: How did you know I hd? I told you this ws rigged! Sdr: Of course it s rigged, or it wouldn t work. Do you wnt to finish or not? Jim: Yeh, I guess so. Go hed, wht do I do net? Sdr: Divide by. Oky, now subtrct the -term. Jim: Just ny old -term? Got ny prticulr coefficient in mind? Sdr: Now stop tesing me. I know you only hve one -term left, so subtrct it. Jim: H, h, I could give you hint bout the coefficient, but tht wouldn t be fir, would it? Sdr: Well you could, nd then I could tell you your number, or you could just tell me the number you hve left fter subtrcting. Jim: Oky, the number I hd left t the end ws 5. Let s see if you cn tell me wht the coefficient of the -term I subtrcted is. Sdr: Ah, then the number you chose t the beginning ws 5, nd the coefficient ws 0! Jim: Hey, you re right! How did you do tht? In your group, follow Sdr s instructions nd determine why she knew Jim s number. Mke up nother set of instructions to use s mgic trick. Be sure to use vribles nd some of the eponent rules or rules for multiplying polynomils tht you lerned in this chpter. Echnge instructions with nother group nd see whether you cn figure out how their trick works.

68 (5-68) Chpter 5 Eponents nd Polynomils C H A P T E R W R A P - U P SUMMARY Definitions Emples Definition of negtive integrl eponents If is nonzero rel number nd n is positive integer, then n. n Definition of zero eponent If is ny nonzero rel number, then 0. The epression 00 is undefined. Rules of Eponents 8 0 Emples If nd b re nonzero rel numbers nd m nd n re integers, then the following rules hold. n Negtive eponent n,, nd n n rules 5 5 5, 5 Find the power nd reciprocl in either order. Product rule m n m n 5 7, 0 7 Quotient rule m n mn 8 5 5, Power of power rule ( m)n mn (5) 56 Power of product rule (b)n nbn () 8 ( ) 6 Power of quotient rule b n n n b Scientific Nottion 9 Emples. Determine the number of plces to move the deciml point by emining the eponent on the 0.. Move to the right for positive eponent nd to the left for negtive eponent. Converting from scientific nottion

69 Chpter 5 Converting to scientific nottion. Count the number of plces (n) tht the deciml point must be moved so tht it will follow the first nonzero digit of the number.. If the originl number ws lrger thn 0, use 0n.. If the originl number ws smller thn, use 0n. Polynomils Summry (5-69) 5 67, Emples Term of polynomil The product of number (coefficient) nd one or more vribles rised to whole number powers, y, 5 Polynomil A single term or finite sum of terms 5 7 Adding or subtrcting polynomils Add or subtrct the like terms. ( ) ( 7) ( ) ( ) Multiplying two polynomils Multiply ech term of the first polynomil by ech term of the second polynomil, then combine like terms. ( )( ) ( ) ( ) 5 Dividing polynomils Ordinry division or long division dividend (quotient)(divisor) reminder dividend reminder quotient divisor divisor Synthetic division A condensed version of long division, used only for dividing by polynomil of the form c If the reminder is 0, then the dividend fctors s 5 ( 7)( ) dividend (quotient)(divisor). Emples FOIL The product of two binomils cn be found quickly by multiplying their First, Outer, Inner, nd Lst terms. ( )( ) 5 6 Squre of sum ( b) b b ( 5) 0 5 Squre of difference ( b) b b (m ) m 6m 9 Product of sum nd difference ( b)( b) b ( )( ) 9 Shortcuts for Multiplying Two Binomils

70 6 (5-70) Chpter 5 Eponents nd Polynomils Fctoring Emples Fctoring polynomil Write polynomil s product of two or more polynomils. A polynomil is fctored completely if it is product of prime polynomils. ( ) ( )( ) Common fctors Fctor out the gretest common fctor (GCF). 6 ( ) Difference of two squres b ( b)( b) (The sum of two squres b is prime.) m 5 (m 5)(m 5) m 5 is prime. Perfect squre trinomils b b ( b) b b ( b) 0 5 ( 5) 6 9 ( ) Difference of two cubes b ( b)( b b ) 8 ( )( ) Sum of two cubes b ( b)( b b) 7 ( )( 9) Grouping Fctor out common fctors from groups of terms. w b bw ( w) b( w) ( b)( w) Fctoring b c By the c method:. Find two numbers tht hve product equl to c nd sum equl to b.. Replce b by two terms using the two new numbers s coefficients.. Fctor the resulting four-term polynomil by grouping. By tril nd error: Try possibilities by considering fctors of the first term nd fctors of the lst term. Check them by FOIL. Substitution Use substitution on higher-degree polynomils to reduce the degree to or. Solving Equtions by Fctoring 9 8 ( )( 9) 8 Let. 8 Emples. Write the eqution with 0 on the right-hnd side.. Fctor the left-hnd side.. Set ech fctor equl to 0.. Solve the simpler equtions. 5. Check the nswers in the originl eqution. Strtegy 7 c 6, b 7, 6 6, ( ) ( ) ( )( ) 8 0 ( 6)( ) or 0 6 or 6 (6) 8 0 () () 8 0

71 Chpter 5 Enriching Your Mthemticl Word Power (5-7) 7 ENRICHING YOUR MATHEMATICAL WORD POWER For ech mthemticl term, choose the correct mening.... polynomil. four or more terms b. mny numbers c. sum of four or more numbers d. single term or finite sum of terms d 0. fctor. to write n epression s product b. to multiply c. wht two numbers hve in common d. to FOIL. degree of polynomil. the number of terms in polynomil b. the highest degree of ny of the terms of polynomil c. the vlue of polynomil when 0 d. the lrgest coefficient of ny of the terms of polynomil b prime number. polynomil tht cnnot be fctored b. number with no divisors c. n integer between nd 0 d. n integer lrger thn tht hs no integrl fctors other thn itself nd d. leding coefficient. the first coefficient b. the lrgest coefficient c. the coefficient of the first term when polynomil is written with decresing eponents d. the most importnt coefficient c gretest common fctor. the lest common multiple b. the lest common denomintor c. the lrgest integer tht is fctor of two or more integers d. the lrgest number in product c. prime polynomil. polynomil tht hs no fctors b. product of prime numbers c. first-degree polynomil d. monomil. fctor completely. to fctor by grouping b. to fctor out prime number c. to write s product of primes d. to fctor by tril-nd-error c. monomil. single polynomil b. one number c. n eqution tht hs only one solution d. polynomil tht hs one term d 5. FOIL. method for dding polynomils b. first, outer, inner, lst c. n eqution with no solution d. polynomil with five terms b 6. binomil. polynomil with two terms b. ny two numbers c. the two coordintes in n ordered pir d. n eqution with two vribles 5. sum of two cubes. ( b) b. b c. b d. b b 7. scientific nottion. the nottion of rtionl eponents b. the nottion of lgebr c. nottion for epressing lrge or smll numbers with powers of 0 d. rdicl nottion c 6. qudrtic eqution. b 0, where 0 b. b c d c. b c 0, where 0 d. ny eqution with four terms c 8. trinomil. polynomil with three terms b. n ordered triple of rel numbers c. sum of three numbers d. product of three numbers 7. zero fctor property. If b 0, then 0 or b 0 b. 0 0 for ny c. 0 for ny rel number d. () 0 for ny rel number synthetic division. division of nonrel numbers b. division by zero c. multipliction tht looks like division d. quick method for dividing by c d difference of two squres. b b. b c. b d. ( b) c

72 8 (5-7) Chpter 5 Eponents nd Polynomils REVIEW EXERCISES 5. Simplify ech epression. Assume ll vribles represent nonzero rel numbers. Write your nswers with positive eponents () 6. () 7. () y5. y (w ) (6w 5) 8. ( y) (5y 7) 5. ( ) ( 7) 6. (7 ) ( 5 6) 7. ( )( ) 8. ( 5)( 0) Perform the indicted opertions. 9. y 7z 5(y z) w. w w m m6. m 5y 6. 5y ( ) 5. m (5m m ) 5. ( ) 5. Perform the following computtions mentlly. Write down only the nswers. 5. ( )( 7) 5. (k 5)(k ) 55. (z 5y)(z 5y) 56. (m )(m ) 57. (m 8) 58. (b ) 59. (w 6)(w ) 60. (w )(w 6) Perform ech computtion without clcultor. Write the nswer in scientific nottion. (,000,000,000)( ) 5. (0.0000)(,000,000) 6. (k ) 6. (n 5) 6. (m 5)(m 5) 6. (k 5t)(k 6t) (. 0)( 05) Find the quotient nd reminder. 65. ( 0) ( ) 5. Simplify ech epression. Assume ll vribles represent 66. ( 5 9) ( ) Write ech number in stndrd nottion Write ech number in scientific nottion.. 8,070, , nonzero rel numbers. Write your nswers with positive eponents. 7. ( ) 7 8. ( y) 67. (m ) (m ) 68. ( ) ( ) 69. (9 8) ( ) 70. ( b) ( b) 9. (mn)(mn) 0. (wy)(wy) b ( b) 7. (b ) y 5 ( ) 8. ( ) Simplify ech epression. Assume tht the vribles represent integers. 9. 5w 5w 5 0. y(y) 5 7 6k k 7. (m 6m 8m) (m) m 7. (w ) ( w), 0 Rewrite ech epression in the form reminder quotient. divisor Use synthetic division

73 Chpter 5 9. Determine whether the first polynomil is fctor of the second. Use synthetic division when possible. 77., Yes 78., 0 Yes 79. 5, 0 Yes 80., 5 No 8., No 8., No 8., 5 6 Yes 8., Yes 5.6 Complete the fctoring by filling in the prentheses ( ) ( ( ) ( ) 88. w w w( ) 89. w w w( ) ( )( ) Fctor ech polynomil. 9. y 8 ( y 9. r t 9v (rt y 0y t 8t w ws s 97. t 5 (t 98. 8y (y 5.7 Fctor ech polynomil ( 00. y y ( y 0. w w 8 (w 0. 6t 5t (t 0. m 5m 7 (m ( 05. m7 m 0m m( 06. 6w5 7w 5w w(w 5.8 Fctor ech polynomil completely ( 08. w 6w 9w w ( 0. ( )(. (. 6 (. y 6y y(. 5m 5 5(m 5. b b b 6. w 6w ( 8. w w (w 0. 8 Review Eercises (5-7) 9 ( (. 6. b b (. 8m m 8 (m ( ) 6 ( ( 7)( 6. (m 6) (m 6) ( )( 8. 5 ( )( 9. ( 9) 5( 9) ( )( Fctor ech polynomil completely. Vribles used s eponents represent positive integers.. k 9. 6k ( k 7)( k 7) ( k )( k k. m m (m )(. yn 7yn 6 (yn )( 5. 9z z (z k ) k k 6. 5z6m 0zm (5z m 7. y by cy bc (y 8. y y ( b b 5.9 Solve ech eqution , 5 0. m 0m 0. ( )( ) 6 0, 5. (w )(w ) 50. m 9m 5 0. m m 9m w 5w w , 8. 7, MISCELL ANEOUS Solve ech problem. 9. Rodrunner nd the coyote. The rodrunner hs just tken position top gint sguro cctus. While positioning 0-foot Acme ldder ginst the cctus, Wile E. Coyote notices wrning lbel on the ldder. For sfety, Acme recommends tht the distnce from the ground to the top of the ldder, mesured verticlly long the cctus, must be feet longer thn the distnce between the bottom of the ldder nd the cctus. How fr from the cctus should he plce the bottom of this ldder?

74 0 (5-7) Chpter 5 Eponents nd Polynomils Socil Security nd the rest from personl svings. To clculte the mount of regulr svings, we use the formul ( i )n S R, i where S is the mount t the end of n yers of n investments of R dollrs ech yer erning interest rte i compounded nnully. ) Use the ccompnying grph to estimte the interest rte needed to get n investment of $ per yer for 0 yers to mount to $00. 5% b) Use the formul to determine the nnul svings for 0 yers tht would mount to $500,000 t 7% compounded nnully. $, Three consecutive integers. Find three consecutive integers such tht the sum of their squres is Perimeter of squre. If the re of squre field is 9 6 squre kilometers, then wht is its perimeter? kilometers 5. Lndscpe design. Rico plnted red tulips in squre flower bed with n re of squre feet (ft). He plns to surround the tulips with dffodils in uniform border with width of feet. Write polynomil tht represents the re plnted in dffodils. 6 ft 5. Life epectncy of blck mles. The ge t which people die is precisely mesured nd provides n indiction of the helth of the popultion s whole. The formul 00 cn be used to model life epectncy L for U.S. blck mles with present ge (Ntionl Center for Helth Sttistics, ) To wht ge cn 0-yer-old blck mle epect to live? 68.7 yers b) How mny more yers is 0-yer-old white mle epected to live thn 0-yer-old blck mle? (See Section 5. Eercise 95.) 6.5 yers Amount (in dollrs) L 6.(.00) L 7.9(.00) cn be used to model life epectncy for U.S. blck femles with present ge. How long cn 0-yer-old blck femle epect to live? 75.9 yers 55. Golden yers. A person erning $80,000 per yer should epect to receive % of her retirement income from Interest rte (percent) FIGURE FOR EXERCISE Life epectncy of blck femles. The formul Amount of sving $ per yer 50 for 0 yers 56. Costly eduction. The verge cost for tuition, room, nd bord for one yer t privte college ws $6, for (Ntionl Center for Eductionl Sttistics, Use the formul in the previous eercise to find the nnul svings for 8 yers tht would mount to $6, with n nnul return of 8%. $.6 CHAPTER 5 TEST Simplify ech epression. Assume ll vribles represent nonzero rel numbers. Eponents in your nswers should be positive eponents y9 5. y y 6. (b) 6 7. (b) 8. 9 Convert to stndrd nottion Perform ech computtion by converting ech number to scientific nottion. Give the nswer in scientific nottion. (80,000)(0.0006).,000,000 ( )(500). (0,000)(0.0) Perform the indicted opertions.. ( 6) ( ). ( 6 7) ( ) 5. ( 7)( ) 5 6. ( 7 7 5) ( ) 7. ( ) ( ) ( )

75 Chpter 5 0. ( 6). ( 5) 0. (y 5)(y 5) Write complete solution for ech problem.. A portble television is dvertised s hving 0-inch digonl mesure screen. If the width of the screen is inches more thn the height, then wht re the dimensions of the screen? Width 8 inches, height 6 inches 5. The infnt mortlity rte for the United Sttes, the number of deths per 00,000 live births, hs decresed drmticlly since 950. The formul d (.8 08)(.0)y gives the infnt mortlity rte d s function of the yer y (Ntionl Center for Helth Sttistics, Find the infnt mortlity rtes in 950, 990, nd , 0.8, 7.9 Fctor completely. 5. ( m (m 8. y y y(. m 7m 5 0 Rewrite ech epression in the form reminder quotient. divisor Use synthetic division (5-75) ( Solve ech eqution. Find the products. 9. ( 7)( ) Test

76 dugoi_c05.qd_c 0/0/00 :9 PM Pge M A K I N G C O N N E C T I O N S C H A P T E R S 5 Simplify ech epression... (.5 0 )w ( 07)(y 5 0) 6 0. ( ) (5)( ) Solve ech problem. 5. Negtive income t. In negtive income t proposl, the function D 0.75E 5000 is used to determine the disposble income D (the mount vilble for spending) for n erned income E (the mount erned). If E D, then the difference is pid in federl tes. If D E, then the difference is pid to the wge erner by Uncle Sm. ) Find the mount of t pid by person who erns $00,000. b) Find the mount received from Uncle Sm by person who erns $0,000. c) The ccompnying grph shows the lines D 0.75E 5000 nd D E. Find the intersection of these lines. d) How much t does person py whose erned income is t the intersection found in prt (c)? () 0.08(68) Solve ech eqution ( 50). 5b c 5c t 5t 0 Disposble income (in thousnds of dollrs) 7. 5u v D = 0.75E ( )( 5) 0 D=E Erned income (in thousnds of dollrs) FIGURE FOR EXERCISE Net Tutor (5-76)