Mth 1051 Dignostic Pretest Ke nd Hoework HW1 The dignostic test is designed to give us n ide of our level of skill in doing high school lgebr s ou begin Mth 1051. You should be ble to do these probles correctl nd without too uch difficult. Even though Mth 1051 begins with brief review of high school lgebr, if ou re rell lost doing these probles ou should tlk to Professor Robertson (61-65-1075, droberts@un.edu) ieditel. Go through the solutions nd be sure ou understnd the. Then, do the hoework probles given in the boxes nd hnd the in t the second lecture session. On the ppers ou turn in for grding, be sure to print our 4-digit hoework ID nuber, our ne, nd HW1 t the top of the first pge. () 4( ) 1. Siplif: 8 6( ) ( ) Follow the order of opertions for siplifing rithetic expressions: Step 1 Siplif expressions inside grouping sbols, which include prentheses ( ), brckets [ ], the frction br, bsolute vlue brs, nd rdicls n. b Step Siplif exponents, squre roots, nd bsolute vlues. Step Siplif ultipliction nd division, working left to right. Step 4 Siplif ddition nd subtrction, working left to right. () 4( ) 6 1 18 8 8 8 8810 6( ) ( ) 1 9 Hoework: Siplif ech of the following. Check the nswers t the end of this docuent. 1. 14 7 1b. 0 4 6 5 1 11 1c. 7 9 6 18 7. Siplif: 1 1 5 (For ore infortion see Sullivn pges A8 - A9.) 10 1 18 Step 1 Find the Lest Coon Denointor (LCD) b prie fctoring ech denointor nd then ultipling the lrgest group of ech prie fctor. 10 = 5 1 = 18 = Lrgest group of 's is (in 1) Lrgest group of 's is (in 18) Lrgest group of 5's is 5 (in 10) So, the LCD = 5 = 180
Mth 1051 Dignostic Pretest Ke nd Hoework HW1 pge Step Convert ech given frction into n equivlent frction with the LCD s the denointor. 1 1 5 10 1 18 1 18 1 15 5 10 10 18 1 15 18 10 18 15 50 180 180 180 Step Now tht the frctions hve the se denointor, dd or subtrct the nuertors. 18 15 50 17 180 180 Step 4 Reduce b cnceling fctors coon to the nuertor nd denointor. 17 17 180 5 Since there re no fctors coon to the nuertor nd denointor the frction cnnot be 17 reduced. The finl nswer is. 180 Hoework: Siplif ech of the following. Check the nswers t the end of this docuent.. 9 1 14 b. 10 9 15 7 5 7 c. 1 1 15 1 10 4 9 6. Siplif: 16 9 50x where x 0, 0 (For ore infortion see Sullivn, pges A8 - A8.) Fctor the rdicnd. Ech pir of identicl fctors is perfect squre nd so tht fctor will coe out fro under the rdicl. An left-over single fctors will rein under the rdicl. 16 9 50x 5 5 x x x x x x x x x x x x x x x x 5 x x x x x x x x 8 4 5x Rther thn writing out ll the fctors of x nd, we could sipl divide their exponents b the index of the rdicl, which is since we hve squre root. The resulting quotient is the exponent of the vrible outside the rdicl nd the resulting reinder is the exponent of the vrible under the rdicl. x 16 8 Exponent of x outside rdicl 16 16 0 No x' s rein under rdicl 9 4 Exponent of outside rdicl 9 8 1 One reins under rdicl Hoework: Siplif ech of the following. Check the nswers t the end of this docuent. 5 7. 7x where x 0, 0 b. 8 7 54x c. 5 9 4 18x
Mth 1051 Dignostic Pretest Ke nd Hoework HW1 pge 4. Fctor: x 5x 1 (For ore infortion see Sullivn pges A8 - A9.) We re going to do LOT of fctoring in this course. Here is generl procedure to follow: Step 1 Fctor out the Gretest Coon Fctor (GCF). 5 4 For exple, fctor: 6x 8x x 5 4 6x 8x x xxx x x xxx x x x xx 1 x x 4 1 The GCF is x. Notice the third ter, 1, ust be there for the fctoriztion to be correct. Step Count the nuber of ters nd look for fctoring ptterns. Two ters: Tr fctoring using one of these ptterns (eorize these!) Difference of perfect squres: x x x Difference or su of perfect cubes: x x x x x xx x Notice the onl difference between the bove two foruls re the signs. Three ters: Tr fctoring using the ptterns for perfect squre trinoils: x x x b b x x x Notice the onl difference between the bove two foruls re the signs. If the trinoil hs the for x bx c, find two integers whose product is c nd whose su is b. Let's s the integers re nd n. Then x bx c x x n. For exple, fctor: x x 1 We need two integers whose product is 1 nd whose su is 1. Integer fctors of 1 re 1 1, 6, nd 4. Since the product is negtive, the two fctors ust hve different signs. Since the su ust be 1, we choose 4. Thus, x x 1 x x 4. If the trinoil hs the for x bx c, find two integers whose product is c nd whose su is b, replce the bx ter using these integers, nd fctor b grouping. For n exple, see below.
Mth 1051 Dignostic Pretest Ke nd Hoework HW1 pge 4 Four ters: Tr fctoring b grouping. For exple, fctor: 5x 15xx 6 Group the first pir of ters nd fctor out the GCF; do the se for the second pir of ters. 5x 15xx 6 5x 15x x 6 5xx x x 5x Step Fctor copletel. In soe cses, it be necessr to fctor ore thn once. Reeber tht ultipliction cn be used to check the fctoriztion. Now, let's fctor x 5x 1: Step 1 GCF: There is no fctor coon to ll the ters so there is no GCF other thn 1 or 1. Step Nuber of ters: There re three ters nd the tch the pttern x bx c, where =, b = 5, nd c = 1. So, we find two integers whose product is c = ( 1) = 4 nd whose su is b = 5. Here re the possibilities for 4 (ignore the sign for the oent): 4 = 1 4, 1, 8, 4 6 The pir tht cn hve su of 5 is 8 if we ttch negtive sign to the. The integers we seek re nd 8 since 8 4 nd 8 5. Replce the iddle ter, 5x, with its equivlent x + 8x nd fctor b grouping: x 5x1x x+8x1 x x 8x 1 xx 4x x x 4 Step Fctor copletel: Since the expression cnnot be further fctored the nswer is x x 4 Hoework: Fctor ech of the following. Check the nswers t the end of this docuent. 4. 4x 64x 56 4b. 6x x 1 4c. 4x 0x 5.
Mth 1051 Dignostic Pretest Ke nd Hoework HW1 pge 5 x 5. Siplif: 9x 5 where x 0, 0 (For ore infortion see Sullivn, pges A8 - A9) Be sure ou hve eorized the lws of exponents: n n n where0 n n n b 0 b 1 where0 1 where0 It usull is esiest to siplif the nuertor nd denointor individull nd then siplif the quotient. 5 x 1 5 6 5 6 9 x 6 5 x x 9 1 9x 9x x x x 6 9 x 6 5 6 8 11 Hoework: Siplif ech of the following. Assue neither x nor re 0. Check the nswers t the end of this docuent. 1 x 6x 4 x x 5b. where x 0, 0 5 10x 5c. 5 6x 5. 4 where x 0, 0
Mth 1051 Dignostic Pretest Ke nd Hoework HW1 pge 6 x x 1 5x 16 6. Solve: (For ore infortion see Sullivn pges A44 - A45.) 6 1 When solving n eqution tht contins frctions, it usull is esiest if ou cler the frctions first b ultipling ech ter b the LCD of ll the ters. The LCD of, 6, nd 1 is 1, so ultipl ech ter b 1 nd then siplif. x x1 5x16 6 1 x x1 5x 16 1 1 1 6 1 4 x x 1 1 5x 16 1 1 1 6 1 4xx1 5x 16 4x6x5x 16 x5x 16 7x 14 x Hoework: Solve ech of the following. Check the nswers t the end of this docuent. x1 6 x 7x 11 6b. x x 6c. x 6. 1 1 1 4 x 1 7 x 6 8 1 Answers to probles: 1. 6 1b. 1c. 11. 1 18. 1 6x x b. b. 1 0 c. 1 9 x x c. 6 5 4 4 x 4x 4. 4x8 x 8 4b. x 4x 4c. x5 x 5 5. 14 7x 5b. 15 18 5x 5c. x 6. x 6b. x 1 6c. x 0