When two numbers are written as the product of their prime factors, they are in factored form.

Transcription

1 10 1 Study Guide Pages Factos Because , we say that 3 and 4 ae factos of 12. In othe wods, factos ae the numbes you multiply to get a poduct. Since , 2 and 6 ae also factos of 12. The only factos of 5 ae 1 and 5. Numbes like 5 that have have exactly two factos, the numbe itself and 1, ae called pime numbes. Numbes like 12 that have moe than two factos ae called composite numbes. Pime Numbe Factos 1, 2 1, 3 1, 29 Poducts of Factos Composite Numbe Factos 1, 2, 4 1, 2, 3, 4, 6, 12 1, 3, 11, 33 Poducts of Factos When two numbes ae witten as the poduct of thei pime factos, they ae in factoed fom. Example 1: Example 2: Wite 45 in factoed fom Keep factoing until all factos ae pime numbes. The factoed fom of 45 is Wite 12x 2 y in factoed fom. 12x 2 y 3 4 x x y x x y The factoed fom of 12x 2 y is x x y. Find the factos of each numbe. Then classify each numbe as pime o composite Facto each monomial a xy 11. c y 2 z Glencoe/McGaw-Hill 60 Algeba: Concepts and Applications

2 10 1 Pactice Pages Factos Find the factos of each numbe. Then classify each numbe as pime o composite Facto each monomial m 2 n x 2 y ab s x 3 yz pq 2 2 Find the GCF of each set of numbes o monomials , , , , 8, , 36, , 36, x, 9x 22. 5y 2, 15y c 2, 13d mn 2, 20m ab 2, 18ab x 2 y 3, 21xy xy, 18y c 2 d, 27cd m, mn Glencoe/McGaw-Hill 60 Algeba: Concepts and Applications

3 10 1 Enichment Pages Finding the GCF by Euclid s Algoithm Finding the geatest common facto of two lage numbes can take a long time using pime factoizations. This method can be avoided by using Euclid s Algoithm as shown in the following example. Example: Find the GCF of 209 and 532. Divide the geate numbe, 532, by the lesse, 209. Divide the emainde into the diviso above. Repeat this pocess until the emainde is zeo. The last nonzeo emainde is the GCF. The diviso, 19, is the GCF of 209 and Suppose the GCF of two numbes is found to be 1. Then the numbes ae said to be elatively pime. Find the GCF of each goup of numbes by using Euclid s Algoithm ; ; ; ; ; ; ,583; ; ; 209; ; 407; a 2 x 2 z; 1615axz cf 3 ; 893c 3 f s 2 ; 495s 3, s x 5 y 7 ; 328xy; 568x 3 y 3 Glencoe/McGaw-Hill 60 Algeba: Concepts and Applications

4 10 2 Study Guide Pages Factoing Using the Distibutive Popety When you use the Distibutive Popety to multiply a monomial by a polynomial, you show two factos and a poduct. Distibutive Popety Factos Poduct 3a(5a 4) 15a 2 12a 3a and 5a 4 15a 2 12a 2x(x 2 6x 1) 2x 3 12x 2 2x 2x and x 2 6x 1 2x 3 12x 2 2x 5s(4 2s) 20 2 s 10s 2 5s and 4 2s 20 2 s 10s 2 When you evese the Distibutive Popety to identify the factos of the poduct, the polynomial is said to be in factoed fom. This is called factoing the polynomial. Example: Facto 15ab 2 12a 2 b 2. 15ab a b b Begin by factoing each monomial. 12a 2 b a a b b Then identify the factos both monomials have in common. The common factos of 15ab 2 and 12a 2 b 2 ae 3 a b b, so the geatest common facto is 3ab 2. 15ab 2 3ab 2 (5) Now wite each monomial as a poduct 12a 2 b 2 3ab 2 (4a) of 3ab 2 and its othe factos. The factoed fom fo 15ab 2 12a 2 b 2 is 3ab 2 (5 4a). Use the Distibutive Popety to check that the factoed fom is equivalent to the given polynomial. Check: 3ab 2 (5 4a) 3ab 2 (5) 3ab 2 (4a) o 15ab 2 12a 2 b 2 Facto each polynomial. If the polynomial cannot be factoed, wite pime a 3b 2. 8w d 2 18d 4. 5c 4 2c mn 3 5n 6. 4g 13h x xy 3 7x 2 y 9. 27pw 25q c 2 d 2 36c 2 d 11. ad 3x gh 2 7g 14gh a 2 15a 20ab 2 Glencoe/McGaw-Hill 61 Algeba: Concepts and Applications

5 10 2 Pactice Pages Factoing Using the Distibutive Popety Facto each polynomial. If the polynomial cannot be factoed, wite pime. 1. 4x y 2 12y 3. 10x 5x 2 y 4. 7yz 3x s 6. 14ab 21a 7. 9xy 3xy m 2 n 18mn ab 2a 2 b a 2 bc 36ab x 2 y 25m x 2 y 3 10xy 13. 4xy 2 18xy 14y 14. 7m 2 28mn 14n x 2 y 4xy 2xy a 3 b 9a 2 b 15b a 2 bc 24ac 2 36a 3 c 18. 8x 3 y 2 16xy 28x 2 y 3 Find each quotient. 19. (6m 2 4) (14x 2 21x) 7x 21. (10x 2 15y 2 ) (2c 2 4c) 2c 23. (12xy 9y) 3y 24. (9a 2 b 27ab) 9ab 25. (25m 2 n 2 15mn) 5mn 26. (3a 2 b 9abc 2 ) 3ab Glencoe/McGaw-Hill 61 Algeba: Concepts and Applications

6 10 2 Enichment Pages Puzzling Pimes A pime numbe has only two factos, itself and 1. The numbe 6 is not pime because it has 2 and 3 as factos; 5 and 7 ae pime. The numbe 1 is not consideed to be pime. 1. Use a calculato to help you find the 25 pime numbes less than , 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 Pime numbes have inteested mathematicians fo centuies. They have tied to find expessions that will give all the pime numbes, o only pime numbes. In the 1700s, Eule discoveed that the expession x 2 x 41 will yield pime numbes fo values of x fom 0 though Find the pime numbes geneated by Eule s fomula fo x fom 0 though 7. 41, 43, 47, 53, 61, 71, 83, Show that the expession x 2 x 31 will not give pime numbes fo vey many values of x. It woks fo x 0, 2, 3, 5, and 6 but not fo x 1, 4, and Find the lagest pime numbe geneated by Eule s fomula Goldbach s Conjectue is that evey nonzeo even numbe geate than 2 can be witten as the sum of two pimes. No one has eve poved that this is always tue. No one has dispoved it, eithe. 5. Show that Goldbach s Conjectue is tue fo the fist 5 even numbes geate than , 6 3 3, 8 3 5, , Give a way that someone could dispove Goldbach s Conjectue. Find an even numbe that cannot be witten as the sum of two pimes. Glencoe/McGaw-Hill 61 Algeba: Concepts and Applications

7 10 3 Study Guide Pages Factoing Tinomials: x 2 bx c To find the two binomial factos of a polynomial, use the FOIL method. Example 1: Facto x 2 5x 6. The fist tem in the tinomial is x 2. Since x x x 2, the fist tem of each binomial is x. x 2 5x 6 (x )(x ) To find the last tems, find a numbe pai whose poduct is 6 and whose sum is 5. Theefoe, x 2 5x 6 (x 2)(x 3). Poduct Factos Sum 6 1, , Example 2: Facto x 2 8x 12. The fist tems ae both x. To find the last tems, find a numbe pai whose poduct is 12 and whose sum is 8. Poduct Factos Sum 12 1, 12 1 ( 12) , 6 2 ( 6) , 4 Once the coect sum is found, it is not necessay to check any moe factos. Theefoe, x 2 8x 12 (x 2)(x 6). Example 3: Facto x 2 2x 15. The fist tems ae both x. To find the last tems, find a numbe pai whose poduct is 15 and whose sum is 2. Poduct Factos Sum 15 1, 15 1 ( 15) , , 5 3 ( 5) 2 Theefoe, x 2 2x 15 (x 3)(x 5). Facto each tinomial. 1. x 2 3x 2 2. w 2 6w z 2 6z 5 5. f 2 6f 8 6. x 2 15x v 2 15v k 2 23k y 2 20y a 2 4a x 2 7x m 2 21m 22 Glencoe/McGaw-Hill 62 Algeba: Concepts and Applications

8 10 3 Pactice Pages Factoing Tinomials: x 2 bx c Facto each tinomial. If the tinomial cannot be factoed, wite pime. 1. x 2 5x 6 2. y 2 5y 4 3. m 2 12m p 2 8p a 2 8a n 2 4n 4 7. x 2 9x x 2 x 3 9. y 2 6y c 2 8c m 2 2m b 2 9b x 2 8x n 2 5n y 2 8y c 2 4c x 2 x m 2 5m a 2 4a y 2 y b 2 3b x 2 3x c 2 2c x 2 10x y 2 15y m 2 10m b 2 6b n 2 12n x 2 8x y 2 15y 12 Glencoe/McGaw-Hill 62 Algeba: Concepts and Applications

9 10 3 Enichment Pages Aea Models fo Quadatic Tinomials Afte you have factoed a quadatic tinomial, you can use the factos to daw geometic models of the tinomial. x 2 5x 6 (x 1)(x 6) To daw a ectangula model, the value 2 was used fo x so that the shote side would have a length of 1. Then the dawing was done in centimetes. So, the aea of the ectangle is x 2 5x 6. x 6 x 1 To daw a ight tiangle model, ecall that the aea of a tiangle is one-half the base times the height. So, one of the sides must be twice as long as the shote side of the ectangula model. x 2 5x 6 (x 1)(x 6) 1 (2x 2)(x 6) 2 The aea of the ight tiangle is also x 2 5x 6. 2x 2 x 6 Facto each tinomial. Then follow the diections to daw each model of the tinomial. 1. x 2 2x x 2 5x 2 Use x 2. Daw a ectangle Use x 1. Daw a ectangle in centimetes. in centimetes. 3. x 2 4x 3 Use x 4. Daw two diffeent ight tiangles in centimetes. 4. 9x 2 9x 2 Use x 2. Daw two diffeent ight tiangles. Use 0.5 centimete fo each unit. Glencoe/McGaw-Hill 62 Algeba: Concepts and Applications

10 10 4 Study Guide Pages Factoing Tinomials: ax 2 bx c To find the two binomial factos of a polynomial, use the FOIL method. Example 1: Facto 5x 2 37x 14. The fist tem in the tinomial is 5x 2. The only factos of 5 ae 5 and 1, so the fist tems of the binomials ae 5x and x. 5x 2 37x 14 (5x )(x ) The last tem in the tinomial is 14, which has two pais of factos, 1 and 14, and 2 and 7. Ty the facto pais until you find the one that gives a middle tem of 37x. Fist Tems Last Tems Binomial Pai Middle Tem Tinomial 5x, x 1, 14 (5x 1)(x 14) x 70x 71x 5x 2 71x 14 5x, x 14, 1 (5x 14)(x 1) 14x 5x 19x 5x 2 19x 14 5x, x 2, 7 (5x 2)(x 7) 2x 35x 37x 5x 2 37x 14 Theefoe, 5x 2 37x 14 (5x 2)(x 7). Example 2: Facto 6x 2 23x 7. Thee ae two possible facto pais of the fist tem, 2x and 3x, and 6x and x. The last tem is positive. The sum of the inside and outside tems is negative. So, the factos of 7 ae 1 and 7. Ty the facto pais until you find the one that gives a middle tem of 23x. Fist Tems Last Tems Binomial Pai Middle Tem Tinomial 2x, 3x 1, 7 (2x 1)(3x 7) 3x 14x 17x 6x 2 17x 7 3x, 2x 1, 7 (3x 1)(2x 7) 2x 21x 23x 6x 2 23x 7 Theefoe, 6x 2 23x 7 (3x 1)(2x 7). Facto each tinomial. 1. 3x 2 4x w 2 3w z 2 14z f 2 27f x 2 3x v 2 10v k 2 9k y 2 3y a 2 6a 8 Glencoe/McGaw-Hill 63 Algeba: Concepts and Applications

11 10 4 Pactice Pages Factoing Tinomials: ax 2 bx c Facto each tinomial. If the tinomial cannot be factoed, wite pime. 1. 2y 2 8y x 2 5x a 2 4a m 2 4m c 2 6c q 2 2q x 2 13x y 2 14y b 2 b a 2 8a n 2 7n x 2 3x c 2 3c y 2 17y b 2 2b x 2 10x m 2 19m a 2 10a b 2 16b y 2 y c 2 11c x 2 x m 2 11m y 2 3y b 2 12b x 2 8x n 2 14n x 2 18x a 2 18a y 2 15y 6 Glencoe/McGaw-Hill 63 Algeba: Concepts and Applications

12 10 4 Enichment Pages Factoing Tinomials of Fouth Degee Some tinomials of the fom a 4 a 2 b 2 b 4 can be witten as the diffeence of two squaes and then factoed. Example: Facto 4x 4 37x 2 y 2 9y 4. Step 1 Find the squae oots of the fist and last tems. 4x 4 2x 2 9y 4 3y 2 Step 2 Find twice the poduct of the squae oots. 2(2x 2 )(3y 2 ) 12x 2 y 2 Step 3 Sepaate the middle tem into two pats. One pat is eithe you answe to Step 2 o its opposite. The othe pat should be the opposite of a pefect squae. 37x 2 y 2 12x 2 y 2 25x 2 y 2 Step 4 Rewite the tinomial as the diffeence of two squaes and then facto. Facto each tinomial. 4x 4 37x 2 y 2 9y 4 (4x 4 12x 2 y 2 9y 4 ) 25x 2 y 2 (2x 2 3y 2 ) 2 25x 2 y 2 [(2x 2 3y 2 ) 5xy][(2x 2 3y 2 ) 5xy] (2x 2 5xy 3y 2 )(2x 2 5xy 3y 2 ) 1. x 4 x 2 y 2 y 4 2. x 4 x a 4 15a a 4 17a a 4 13a a 4 26a 2 b 2 25b x 4 21x 2 y 2 9y a 4 29a 2 c 2 25c 4 Glencoe/McGaw-Hill 63 Algeba: Concepts and Applications

13 10 5 Study Guide Pages Special Factos x 2 10x 25 (x 5)(x 5) Tinomials like the one above that have two equal binomial factos ae called pefect squae tinomials. Recall that when a numbe is multiplied by itself, the esult is a pefect squae. Fo example, , so 16 is a pefect squae. Factoing Pefect Symbols: a 2 2ab b 2 (a b)(a b) Squae Tinomials a 2 2ab b 2 (a b)(a b) Example: x 2 2x 1 (x 1)(x 1) Studying these popeties of x 2 10x 25 will help you facto othe pefect squae tinomials. 1. The fist tem, x 2, is a pefect squae. x x x 2 2. The last tem, 25, is a pefect squae The middle tem, 10x, is twice the poduct of 5 and x. 2(5x) 10x Example 1: Facto y 2 14y 49. The fist tem is a pefect squae. The last tem is a pefect squae. The middle tem is twice the poduct of the fist and last tems. So y 2 14y 49 is a pefect squae tinomial. y 2 14y 49 (y 7)(y 7) o (y 7) 2 When two pefect squaes ae subtacted, the polynomial is called the diffeence of two squaes. The diffeence of two squaes also has a special pai of factos. Factoing Diffeences Symbols: a 2 b 2 (a b)(a b) of Squaes Example: x 2 25 (x 5)(x 5) Example 2: Facto g g 2 36 is the diffeence of two squaes. g 2 36 (g 6)(g 6) Facto each pefect squae tinomial. 1. y 2 16y a 2 14a s 2 10s p 2 20p h 2 12h a 2 12a v 2 24v 16 Facto each diffeence of squaes. 9. b k x m y d 2 Glencoe/McGaw-Hill 64 Algeba: Concepts and Applications

14 10 5 Pactice Pages Special Factos Detemine whethe each tinomial is a pefect squae tinomial. If so, facto it. 1. y 2 6y 9 2. x 2 4x 4 3. n 2 6n 3 4. m 2 12m y 2 10y a 2 16a x 2 6x n 2 20n y 2 9y 9 Detemine whethe each binomial is the diffeence of squaes. If so, facto it. 10. x b y m a n x 2 12y m 2 18n 2 Facto each polynomial. If the polynomial cannot be factoed, wite pime a x x 2 5x y 2 6y m 2 9n a 2 8a b 2 10b 26. 9y 2 12y x 2 3x 18 Glencoe/McGaw-Hill 64 Algeba: Concepts and Applications

15 10 5 Enichment Pages Witing Expessions of Aea in Factoed Fom Wite an expession in factoed fom fo the aea A of the shaded egion in each figue below. 1. x 2. x y x y y y x b a a b a 2a R x 4x 2x Glencoe/McGaw-Hill 64 Algeba: Concepts and Applications

16 10 2 School-to-Wokplace Pages Paking Lot Dimensions (Enginee) Pat of the successful design of any office complex is the planning of adequate paking space fo employees and visitos. Paking engineeing technicians help achitects, developes, and buildes accomplish this task. As you might expect, the amount of paking space is vaiable and depends on the employee capacity of the building. The diagam at the ight shows an office building with space adjacent to it designated fo paking spaces. One dimension of each paking lot is labeled x because the paking technician may want to expeiment with diffeent dimensions. Suppose the technician has detemined that one vehicle needs an 84-squae-foot space. Find an expession fo the numbe of cas the lots can accommodate at one time. Find the aea of each lot. Paking Lot A: x(120 x) Paking Lot B: (60)x The total aea is x(120 x) 60x. Facto using the Distibutive Popety. The lots can hold Solve. x(120 x) 60x x(120 x 60) x(180 x) 84 x(180 x) cas at one time. 1. Will a choice of x 40 feet make it possible to accommodate 200 cas? x 120 ft 60 ft building 120 ft A open space 60 ft B x 2. Expeiment to find the smallest value of x that will make it possible to accommodate 200 cas. 500 ft 3. If the technician detemines that one ca needs an 84-squae-foot space, find an expession fo the numbe of cas the lots epesented at the ight can accommodate at one time. 400 ft 500 ft B A x x Glencoe/McGaw-Hill 10 Algeba: Concepts and Applications