REVIEW OF ALGEBRA FRACTIONS To dd two frctions with the sme denomintor, we use the Distributive Lw: Thus, it is true tht b c b b b c c c b b c b b c b

REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b b b b c b c b c b c bc bc (Commuttive Lw) (Associtive Lw) (Distributive lw) In prticulr, putting in the Distributive Lw, we get b c b c b c nd so b c b c EXAMPLE () 3y4 34 y y (b) t7 t 4t 4t t (c) 4 3 4 3 6 0 3 If we use the Distributive Lw three times, we get bc d bc bd c bc d bd This sys tht we multiply two fctors by multiplying ech term in one fctor by ech term in the other fctor nd dding the products. Schemticlly, we hve bc d In the cse where c nd d b, we hve Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. or Similrly, we obtin EXAMPLE () (b) (c) b b b b b b b b b b 3 5 6 3 0 5 6 7 5 6 36 3 4 3 6 34 3 3 9 5

REVIEW OF ALGEBRA FRACTIONS To dd two frctions with the sme denomintor, we use the Distributive Lw: Thus, it is true tht b c b b b c c c b b c b b c b But remember to void the following common error: b c b c (For instnce, tke b c to see the error.) To dd two frctions with different denomintors, we use common denomintor: b c d bc d bd We multiply such frctions s follows: b c d c bd In prticulr, it is true tht b b b To divide two frctions, we invert nd multiply: EXAMPLE 3 3 () 3 3 (b) 3 3 3 6 (c) s t u ut s t u u s t b c d b d c d bc 6 Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 3 (d) y y y y y y y y y y y y y y FACTORING We hve used the Distributive Lw to epnd certin lgebric epressions. We sometimes need to reverse this process (gin using the Distributive Lw) by fctoring n epression s product of simpler ones. The esiest sitution occurs when the epression hs common fctor s follows: Epnding 3(-)3@-6 Fctoring To fctor qudrtic of the form b c we note tht r s r s rs so we need to choose numbers r nd s so tht r s b nd rs c. EXAMPLE 4 Fctor 5 4. SOLUTION The two integers tht dd to give 5 nd multiply to give 4 re 3 nd 8. Therefore 5 4 3 8 EXAMPLE 5 Fctor 7 4. SOLUTION Even though the coefficient of is not, we cn still look for fctors of the form r nd s, where rs 4. Eperimenttion revels tht 7 4 4 Some specil qudrtics cn be fctored by using Equtions or (from right to left) or by using the formul for difference of squres: Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. 3 The nlogous formul for difference of cubes is 4 which you cn verify by epnding the right side. For sum of cubes we hve 5 b b b 3 b 3 b b b 3 b 3 b b b EXAMPLE 6 () 6 9 3 (Eqution ;, b 3) (b) 4 5 5 5 (Eqution 3;, b 5) (c) 3 8 4 (Eqution 5;, b )

4 REVIEW OF ALGEBRA 6 EXAMPLE 7 Simplify. 8 SOLUTION Fctoring numertor nd denomintor, we hve 6 4 4 8 4 4 To fctor polynomils of degree 3 or more, we sometimes use the following fct. 6 The Fctor Theorem If P is polynomil nd Pb 0, then b is fctor of P. EXAMPLE 8 Fctor 3 3 0 4. SOLUTION Let P 3 3 0 4. If Pb 0, where b is n integer, then b is fctor of 4. Thus, the possibilities for b re,, 3, 4, 6, 8,, nd 4. We find tht P, P 30, P 0. By the Fctor Theorem, is fctor. Insted of substituting further, we use long division s follows: 3 3 0 4 3 0 4 4 Therefore 3 3 0 4 3 4 COMPLETING THE SQUARE Completing the squre is useful technique for grphing prbols or integrting rtionl functions. Completing the squre mens rewriting qudrtic b c in the form p q nd cn be ccomplished by:. Fctoring the number from the terms involving.. Adding nd subtrcting the squre of hlf the coefficient of. In generl, we hve EXAMPLE 9 Rewrite by completing the squre. SOLUTION b c b c b b b c 4 b The squre of hlf the coefficient of is. Thus 4 4 ( ) 3 4 4 b c Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 5 EXAMPLE 0 6 6 9 9 3 9 3 7 QUADRATIC FORMULA By completing the squre s bove we cn obtin the following formul for the roots of qudrtic eqution. 7 The Qudrtic Formul The roots of the qudrtic eqution b c 0 re b sb 4c EXAMPLE Solve the eqution 5 3 3 0. SOLUTION With 5, b 3, c 3, the qudrtic formul gives the solutions 3 s3 453 5 3 s69 0 The quntity b 4c tht ppers in the qudrtic formul is clled the discriminnt. There re three possibilities:. If b 4c 0, the eqution hs two rel roots.. If b 4c 0, the roots re equl. 3. If b 4c 0, the eqution hs no rel root. (The roots re comple.) These three cses correspond to the fct tht the number of times the prbol y b c crosses the -is is,, or 0 (see Figure ). In cse (3) the qudrtic b c cn t be fctored nd is clled irreducible. y y y Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. FIGURE Possible grphs of y@+b+c 0 0 0 () b@-4c>0 (b) b@-4c0 (c) b@-4c<0 EXAMPLE The qudrtic is irreducible becuse its discriminnt is negtive: b 4c 4 7 0 Therefore, it is impossible to fctor.

6 REVIEW OF ALGEBRA THE BINOMIAL THEOREM Recll the binomil epression from Eqution : b b b If we multiply both sides by b nd simplify, we get the binomil epnsion 8 b 3 3 3 b 3b b 3 Repeting this procedure, we get In generl, we hve the following formul. b 4 4 4 3 b 6 b 4b 3 b 4 9 The Binomil Theorem If k is positive integer, then b k k k k b kk kk k 3 kb k b k k b k3 b 3 kk k n 3 n kn b n EXAMPLE 3 Epnd 5. SOLUTION Using the Binomil Theorem with, b, k 5, we hve 5 5 5 4 5 4 3 5 4 3 3 3 5 4 5 5 0 4 40 3 80 80 3 RADICALS The most commonly occurring rdicls re squre roots. The symbol s mens the positive squre root of. Thus s mens Since 0, the symbol s mkes sense only when 0. Here re two rules for working with squre roots: s 0 sb s sb b sb However, there is no similr rule for the squre root of sum. In fct, you should remember to void the following common error: s b s sb (For instnce, tke 9 nd b 6 to see the error.) nd 0 Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 7 EXAMPLE 4 () s8 s 8 s9 3 Notice tht becuse s indictes the positive squre root. (See Absolute Vlue.) s In generl, if n is positive integer, (b) s y s sy sy s n mens n If n is even, then 0 nd 0. Thus s 3 8 becuse 3 8, but s 4 8 nd s 6 8 re not defined. The following rules re vlid: s n s n b s n s n b b s n b EXAMPLE 5 s 3 4 s 3 3 s 3 3 s 3 s 3 To rtionlize numertor or denomintor tht contins n epression such s s sb, we multiply both the numertor nd the denomintor by the conjugte rdicl s sb. Then we cn tke dvntge of the formul for difference of squres: (s sb)(s sb) (s) (sb) b s 4 EXAMPLE 6 Rtionlize the numertor in the epression. SOLUTION We multiply the numertor nd the denomintor by the conjugte rdicl s 4 : s 4 s 4 s 4 s 4 (s 4 ) s 4 4 4 (s 4 ) Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. EXPONENTS Let be ny positive number nd let n be positive integer. Then, by definition,.. 3. 4. n n fctors 0 n n n s n mn s n m (s n ) m m is ny integer

8 REVIEW OF ALGEBRA Lws of Eponents Let nd b be positive numbers nd let r nd s be ny rtionl numbers (tht is, rtios of integers). Then. r s rs. 3. rs s 4. br r b r 5. r b r r b b 0 r r s rs In words, these five lws cn be stted s follows:. To multiply two powers of the sme number, we dd the eponents.. To divide two powers of the sme number, we subtrct the eponents. 3. To rise power to new power, we multiply the eponents. 4. To rise product to power, we rise ech fctor to the power. 5. To rise quotient to power, we rise both numertor nd denomintor to the power. EXAMPLE 7 () 8 8 8 3 8 6 4 (b) y y y y y y yy y y y y y y y y y y (c) 4 3 s4 3 s64 8 Alterntive solution: 4 3 (s4) 3 3 8 (d) (e) s 3 4 y 3 y 43 43 4 3 z y y 8 4 3 z 4 7 y 5 z 4 INEQUALITIES When working with inequlities, note the following rules. Rules for Inequlities. If b, then c b c.. If b nd c d, then c b d. 3. If b nd c 0, then c bc. 4. If b nd c 0, then c bc. 5. If 0 b, then b. Rule sys tht we cn dd ny number to both sides of n inequlity, nd Rule sys tht two inequlities cn be dded. However, we hve to be creful with multipliction. Rule 3 sys tht we cn multiply both sides of n inequlity by positive number, but Rule 4 sys tht if we multiply both sides of n inequlity by negtive number, then we reverse the direction of the inequlity. For emple, if we tke the inequlity Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 9 3 5 nd multiply by, we get 6 0, but if we multiply by, we get 6 0. Finlly, Rule 5 sys tht if we tke reciprocls, then we reverse the direction of n inequlity (provided the numbers re positive). EXAMPLE 8 Solve the inequlity 7 5. SOLUTION The given inequlity is stisfied by some vlues of but not by others. To solve n inequlity mens to determine the set of numbers for which the inequlity is true. This is clled the solution set. First we subtrct from ech side of the inequlity (using Rule with c ): 7 4 Then we subtrct 7 from both sides (Rule with c 7): 6 4 Now we divide both sides by 6 (Rule 4 with c 6): These steps cn ll be reversed, so the solution set consists of ll numbers greter thn. In other words, the solution of the inequlity is the intervl ( 3, ). 3 4 6 3 EXAMPLE 9 Solve the inequlity 5 6 0. SOLUTION First we fctor the left side: 3 0 We know tht the corresponding eqution 3 0 hs the solutions nd 3. The numbers nd 3 divide the rel line into three intervls:,, 3 3, On ech of these intervls we determine the signs of the fctors. For instnce,,?? 0 Then we record these signs in the following chrt: Intervl 3 3 Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. A visul method for solving Emple 9 is to use grphing device to grph the prbol y 5 6 (s in Figure ) nd observe tht the curve lies on or below the -is when 3. y y -5+6 0 3 4 FIGURE 3 3 Another method for obtining the informtion in the chrt is to use test vlues. For instnce, if we use the test vlue for the intervl,, then substitution in 5 6 gives 5 6 The polynomil 5 6 doesn t chnge sign inside ny of the three intervls, so we conclude tht it is positive on,. Then we red from the chrt tht 3 is negtive when 3. Thus, the solution of the inequlity 3 0 is 3, 3

0 REVIEW OF ALGEBRA 0 3 FIGURE 3 + - + Notice tht we hve included the endpoints nd 3 becuse we re looking for vlues of such tht the product is either negtive or zero. The solution is illustrted in Figure 3. EXAMPLE 0 Solve 3 3 4. SOLUTION First we tke ll nonzero terms to one side of the inequlity sign nd fctor the resulting epression: 3 3 4 0 or 4 0 As in Emple 9 we solve the corresponding eqution 4 0 nd use the solutions 4, 0, nd to divide the rel line into four intervls, 4, 4, 0, 0,, nd,. On ech intervl the product keeps constnt sign s shown in the following chrt. Intervl 4 4 4 4 0 0 _4 FIGURE 4 0 Then we red from the chrt tht the solution set is 4 0 or 4, 0, The solution is illustrted in Figure 4. Remember tht if is negtive, then is positive. ABSOLUTE VALUE The bsolute vlue of number, denoted by, is the distnce from to 0 on the rel number line. Distnces re lwys positive or 0, so we hve For emple, In generl, we hve EXAMPLE Epress SOLUTION 3 3 3 3 0 0 s s 3 3 3 0 if 0 if 0 without using the bsolute-vlue symbol. 3 3 3 3 3 for every number if 3 0 if 3 0 if 3 if 3 Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA Recll tht the symbol s mens the positive squre root of. Thus, sr s mens s r nd s 0. Therefore, the eqution s is not lwys true. It is true only when 0. If 0, then 0, so we hve s. In view of (), we then hve the eqution 3 s which is true for ll vlues of. Hints for the proofs of the following properties re given in the eercises. Properties of Absolute Vlues Suppose nd b re ny rel numbers nd n is n integer. Then.. b b b 0 3. n n b b For solving equtions or inequlities involving bsolute vlues, it s often very helpful to use the following sttements. Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. _ FIGURE 5 b 0 -b -b FIGURE 6 Length of line segment -b FIGURE 7 b 3 5 7 Suppose 0. Then 4. if nd only if 5. if nd only if 6. if nd only if or For instnce, the inequlity sys tht the distnce from to the origin is less thn, nd you cn see from Figure 5 tht this is true if nd only if lies between nd. If nd b re ny rel numbers, then the distnce between nd is the bsolute vlue of the difference, nmely, b, which is lso equl to b b. (See Figure 6.) EXAMPLE Solve 5 3 SOLUTION By Property 4 of bsolute vlues, is equivlent to 5 3 or 5 3 So 8 or. Thus, 4 or. 5 EXAMPLE 3 Solve. 5 3 5 SOLUTION By Property 5 of bsolute vlues, is equivlent to 5 Therefore, dding 5 to ech side, we hve 3 7 nd the solution set is the open intervl 3, 7. SOLUTION Geometriclly, the solution set consists of ll numbers whose distnce from 5 is less thn. From Figure 7 we see tht this is the intervl 3, 7.

REVIEW OF ALGEBRA 3 4 EXAMPLE 4 Solve. 3 4 SOLUTION By Properties 4 nd 6 of bsolute vlues, is equivlent to 3 4 or 3 4 In the first cse, 3, which gives 3. In the second cse, 3 6, which gives. So the solution set is or 3, [ 3, ) 6 Epnd nd simplify. 7. 8. 3. 4. A EXERCISES Click here for nswers.. 6b0.5c. yy 4 3. 5 4. 4 3 5. 4 3 6. 8 4 4 5 53t 4 t tt 3 9. 4 3 7 0... 3 y 4 6 y5 y t 5 t 38t 5. 3 6. 7 8 Perform the indicted opertions nd simplify. 8 7. 8. 9. 0. 5 3. u u. u y 3. 4. z 5. r s 6. s 6t c 7. 8. c 9 48 Fctor the epression. 3 b 4 b yz bc b c 9. 3 30. 5b 8bc 3. 7 6 3. 6 33. 8 34. 7 4 35. 9 36 36. 8 0 3 37. 6 5 6 38. 0 5 S 9b 6 3b Click here for solutions. 39. t 3 40. 4t 9s 4. 4t t 9 4. 3 7 43. 3 44. 3 4 5 45. 3 3 3 46. 3 3 60 47. 3 5 4 48. 3 3 4 49 54 Simplify the epression. 49. 50. 3 5. 5. 9 8 53. 54. 3 9 5 4 55 60 Complete the squre. 55. 5 56. 6 80 57. 5 0 58. 3 59. 4 4 60. 3 4 50 6 68 Solve the eqution. 3 4 3 5 6 6. 9 0 0 6. 8 0 63. 9 0 64. 7 0 65. 3 5 0 66. 7 0 67. 3 0 68. 3 3 0 69 7 Which of the qudrtics re irreducible? 69. 3 4 70. 9 4 7. 3 6 7. 3 6 73 76 Use the Binomil Theorem to epnd the epression. 73. b 6 74. b7 75. 4 76. 3 5 Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 3 Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. 77 8 Simplify the rdicls. s 3 s 4 3 4 77. s3 s 78. 79. s 3 54 s 4 80. sy s 3 y 8. s6 4 b 3 8. 83 00 Use the Lws of Eponents to rewrite nd simplify the epression. 9 4 n n 85. 86. 3 3 b 4 87. 88. 5 b 5 96 89. 3 90. 5 9. 5 3 9. 43 64 93. y 4 3 94. 5 y 3 z 0 35 97. 98. (st) 5 t sst 4 s 3 99. 00. 0 08 Rtionlize the epression. s 3 0. 0. 9 s 8 03. 04. 4 05. 06. 3 s5 09 6 Stte whether or not the eqution is true for ll vlues of the vrible. 09. s 0. 6.. 6 6 3. 4. y y 5. 3 4 7 6. 6 4 6 4 4 7 6 Rewrite the epression without using the bsolute vlue symbol. 5 3 s5 5 7. 8. 9. 0.. if. if 3. 4. 5. 6. 8 s 5 s 4 3 n y y 95. s 96. (s) 5 y 6 4 3 (s) s sy s 5 96 6 s 5 3 83. 3 0 9 8 84. 6 4 0 6 6 s 4 r n s 4 r s h s h h 07. s 3 4 08. s s s 4 y y 4 3 7 4 Solve the inequlity in terms of intervls nd illustrte the solution set on the rel number line. 7. 7 3 8. 4 3 6 9. 30. 5 5 3 3. 0 3. 3 4 6 33. 0 34. 8 35. 3 36. 5 37. 3 0 38 3 0 39. 3 40. 3 3 4 4. 4. 3 4 43. The reltionship between the Celsius nd Fhrenheit temperture scles is given by C 5 9 F 3, where C is the temperture in degrees Celsius nd F is the temperture in degrees Fhrenheit. Wht intervl on the Celsius scle corresponds to the temperture rnge 50 F 95? 44. Use the reltionship between C nd F given in Eercise 43 to find the intervl on the Fhrenheit scle corresponding to the temperture rnge 0 C 30. 45. As dry ir moves upwrd, it epnds nd in so doing cools t rte of bout C for ech 00-m rise, up to bout km. () If the ground temperture is 0C, write formul for the temperture t height h. (b) Wht rnge of temperture cn be epected if plne tkes off nd reches mimum height of 5 km? 46. If bll is thrown upwrd from the top of building 8 ft high with n initil velocity of 6 fts, then the height h bove the ground t seconds lter will be During wht time intervl will the bll be t lest 3 ft bove the ground? 47 48 Solve the eqution for. 3 47. 48. 49 56 Solve the inequlity. 3 4 5 3 0.4 49. 50. 5. 5. 53. 54. 55. 56. 57. Solve the inequlity b c bc for, ssuming tht, b, nd c re positive constnts. 58. Solve the inequlity b c for, ssuming tht, b, nd c re negtive constnts. b b h 8 6t 6t 59 Prove tht. [Hint: Use Eqution 3.] 60. Show tht if 0 b, then b. 3 5 3 6 0. 3 5 6

4 REVIEW OF ALGEBRA S ANSWERS Click here for solutions.. 3 bc. 3 y 5 3. 0 4. 4 3 5. 8 6 6. 4 7. 6 3 8. 3t t 9. 5 7 0. 3. 4 4. 9 4 3. 30y 4 y 5 y 6 4. 5t 56t 3 5. 3 5 6. 4 3 7. 4 8. 3 b 9. 3 7 u 3u 0.. 5 u. b 3b 4 z rs 3. 4. 5. b yz y 3t 6. c 3 7. 8. b c 9. 6 30. b5 8c 3. 6 3. 3 33. 4 34. 4 35. 9 36. 4 3 37. 3 3 38. 5 39. t t t 40. t 3st 3s 4. t 3 4. 3 3 9 43. 44. 45. 3 46. 3 5 4 47. 3 4 48. 3 49. 50. 5. 5. 8 4 53. 6 4 54. 9 4 55. 4 56. 8 6 57. ( 5 ) 5 4 58. ( 3 ) 5 4 59. 3 60. 3 4 6., 0 6., 4 63. 9 s85 5 s3 64. s 65. 6 66. 7 s33 s5 67., 4 68., s 69. Irreducible 70. Not irreducible 7. Not irreducible (two rel roots) 7. Irreducible 73. 74. 6 6 5 b 5 4 b 0 3 b 3 5 b 4 6b 5 b 6 7 7 6 b 5 b 35 4 b 3 35 3 b 4 b 5 7b 6 b 7 75. 76. 8 4 6 6 4 4 43 405 70 4 90 6 5 8 0 77. 8 78. 3 79. 80. y 8. 4 bsb 8. 83. 3 6 84. 60 85. 6 0 86. n3 87. y 88. 89. b y s3 90. 5 s3 9. 5 9. 56 93. s 3 y 6 94. 3 95. y 65 96. 34 97. t 5 98. y 95 z 6 8 t 4 99. 00. r n 0. 0. s 4 s 3 4 6 03. 04. s 8 s h s h 3 s5 s sy 3 4 05. 06. 07. y s 3 4 08. 09. Flse 0. Flse s s. True. Flse 3. Flse 4. Flse 5. Flse 6. True 7. 8 8. 9. 5 s5 0... 3. 4. 5. 6. 7., 8. _ 9., 30. _ 0 0 3. 0, 3., 4 0 33.,, 34., 4 35. (s3, s3) 36. (, s5] [s5, ) _œ 3 0 œ 3 37., 38. 3,, 0 39., 0, 40., 0, 3 _ 0 4. 4. (, 3 ),, 0 ( 4, ) 0 4 if if if if if s s if s or s (, 3] (, ) 3 _œ 5 _3 _ 0 3 _ 0 3 0 4 0 œ 5 0 43. 0, 35 44. 68, 86 45. () T 0 0h, 0 h (b) 30C T 0C 46. 0, 3 47., 4 48. 4 3 3, 49. 3, 3 50., 3 3, 5. 3, 5 5. 5.9, 6. 53., 7 3, 54., 4, 55..3,.7 56. ( 4 5, 8 5) bc 57. 58. c b b 0 0 s 0 4 Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 5 SOLUTIONS Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.. ( 6b)(0.5c) ( 6)(0.5)( bc) 3 bc. ( y)( y 4 ) yy 4 3 y 5 3. ( 5) 5 0 4. (4 3) 4 3 4 3 5. (4 3) 4+ 3 8+6 6. 8 (4 + ) 8 4 4 7. 4( +) 5( +)4 4 +8 5 5( ) 5 4 5 4 +0 +8 5 +6 +3 8. 5(3t 4) (t +) t(t 3) 5t 0 t t +6t ( )t +(5+6)t 0 3t +t 9. (4 )(3 +7)4(3 +7) (3 +7) +8 3 7 +5 7 0. ( )( +)( )( +) ( +) ( +) 3 + 3 +. ( ) () ()() + 4 4 +. ( + 3) +()(3)+(3) 9 + +4 3. y 4 (6 y)(5 + y)y 4 [6(5 + y) y(5 + y)] y 4 (30 + 6y 5y y ) y 4 (30 + y y )30y 4 + y 5 y 6 4. (t 5) (t + 3)(8t ) t (5t)+5 (8t t +4t 3) t 0t +5 6t +t 48t +6 5t 56t +3 5. ( + )( 3 +)( 3 +)+( 3 +) 3 ++ 3 6 + 3 5 + 6. ( + ) (+ )( + )(+ )+( + ) ( + ) 7. 8. 9. 0. + + + 3 3 + 4 4 3 + + +8 + 8 +4 9b 6 9b 3b 3b 6 3b 3 b +5 + ()( 3) + ( +5) 3+ +0 3 ( +5)( 3) ( +5)( 3) 3 +7 + 5 + + ( ) + ( +) + + ( +)( ). u ++ u (u +)(u +)+u u +u ++u u +3u + u + u + u + u +. 3. 6. 3 b + 4 b b b 3b b + 4 b b 3b +4 b /y z /y z/ z y yz 4. y/z / y/z z y z y r s 5. rs s 6t 6st rs 3t bc b c bc c b c b c b

6 REVIEW OF ALGEBRA 7. + c c 8. + + + c + c c c + c c c c + + + c c 9. + 3 + 6 ( + 6 ) 30. 5b 8bc b 5 b 8c b(5 8c) c c c c + + + ++ 3+ + + + 3. Thetwointegersthtddtogive7ndmultiplytogive6re6nd.Therefore +7 +6( +6)( +). 3. The two integers tht dd to give ndmultiplytogive 6 re 3 nd. Therefore 6( 3)( +). 33. The two integers tht dd to give ndmultiplytogive 8 re 4 nd. Therefore 8( 4)( +). 34. +7 4( )( +4) 35. 9 36 9( 4) 9( )( +) [Eqution 3 with, b ] 36. 8 +0 +3(4 + 3)( +) 37. 6 5 6(3 + )( 3) 38. +0 +5( +5) [Eqution with, b 5] 3 39. t +(t +)(t t +) [Eqution 5 with t, b ] 40. 4t 9s (t) (3s) (t 3s)(t +3s) [Eqution 3 with t, b 3s] 4. 4t t +9(t 3) [Eqution with t, b 3] 4. 3 7 ( 3)( +3 +9) [Eqution 4 with, b 3] 43. 3 + + ( + +)( +) [Eqution with, b ] 44. Let p() 3 4 +5, nd notice tht p() 0,sobytheFctorTheorem,( ) is fctor. Use long division (s in Emple 8): 3 + 3 4 +5 3 3 +5 3 +3 Therefore 3 4 +5 ( )( 3 +)( )( )( ) ( ) ( ). 45. Let p() 3 +3 3, nd notice tht p() 0, so by the Fctor Theorem, ( ) is fctor. Use long division (s in Emple 8): +4 + 3 3 +3 3 3 4 4 4 3 3 3 3 Therefore 3 +3 3( )( +4 +3)( )( +)( +3). Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 7 46. Let p() 3 3 +60, nd notice tht p(3) 0,sobytheFctorTheorem,( 3) is fctor. Use long division (s in Emple 8): + 0 3 3 3 +60 3 3 3 3 0 +60 0 +60 Therefore 3 3 +60( 3)( + 0) ( 3)( +5)( 4). 47. Let p() 3 +5 4, nd notice tht p() 3 +5() () 4 0,sobytheFctorTheorem, ( ) is fctor. Use long division (s in Emple 8): +7 + 3 +5 4 3 7 7 4 4 4 Therefore 3 +5 4 ( )( +7 + ) ( )( +3)( +4). 48. Let p() 3 3 4 +, nd notice tht p() 0,sobytheFctorTheorem,( ) is fctor. Use long division (s in Emple 8): 6 3 3 4 + 3 4 + 6 + 6 + Therefore 3 3 4 +( )( 6) ( )( 3)( +). Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. 49. 50. 5. 5. 53. 54. + ( +)( ) 3 + ( )( ) + 3 4 ( +)( ) ( )( +) + + ( )( +) 9 +8 ( 8)( ) + 8 3 +5 +6 ( +5 +6) ( +3)( +) ( +) ( 4)( +3) ( 4)( +3) 4 +3 + 9 +3 + ( 3) + ( 3)( +3) ( 3)( +3) 9 + 5 +4 ( )( +) ( 4) ( +) ( 4)( ) ( )( +)( 4) 4 4 ( )( +)( 4) 6 4 ( )( +)( 4) 55. + +5[ +]+5[ + +() () ]+5( +) +5 ( +) +4

8 REVIEW OF ALGEBRA 56. 6 +80[ 6]+80[ 6 +(8) (8) ]+80( 8) +80 64 ( 8) +6 57. 5 +0[ 5]+0 5 + 5 5 +0 5 +0 5 5 + 5 4 4 58. +3 +[ +3]+ 59. 4 +4 4[ + ] 4 +3 + 3 + + 3 + + 3 + 3 + 3 5 4 4 + 4 4 4 + 3 60. 3 4 +503[ 8]+503[ 8 +( 4) ( 4) ]+503( 4) +50 3( 4) 3( 4) + 6. 9 0 0 ( +0)( ) 0 +00or 0 0 or. 6. 80 ( 4)( +)0 40or +0 4or. 63. Using the qudrtic formul, +9 0 9 ± 9 4()( ) () 64. Using the qudrtic formul, 70 ± 4 4()( 7) 65. Using the qudrtic formul, 3 +5 +0 5 ± 5 4(3)() (3) 66. Using the qudrtic formul, +7 +0 7 ± 49 4()() () 9 ± 85. ± 3 5 ± 3. 6 7 ± 33. 4 67. Let p() 3 +, nd notice tht p() 0, so by the Fctor Theorem, ( ) is fctor. Use long division: + 3 +0 + 3 + + ±. Therefore 3 +( )( + ) 0 0 or + 0 or [using the qudrtic formul] ± 4()( ) () ± 5. 68. Let p() 3 +3 +, nd notice tht p( ) 0, so by the Fctor Theorem, ( +)is fctor. Use long division: + + 3 +3 + 3 + + + Therefore 3 +3 + ( +)( + ) 0 +0or + 0 or [using the qudrtic formul] ± 4()( ) ±. Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 9 Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. 69. +3 +4is irreducible becuse its discriminnt is negtive: b 4c 9 4()(4) 3 < 0. 70. The qudrtic +9 +4is not irreducible becuse b 4c 9 4()(4) 49 > 0. 7. 3 + 6 is not irreducible becuse its discriminnt is nonnegtive: b 4c 4(3)( 6) 73 > 0. 7. The qudrtic +3 +6is irreducible becuse b 4c 3 4()(6) 5 < 0. 73. Using the Binomil Theorem with k 6we hve ( + b) 6 6 +6 5 b + 6 5 4 b + 6 5 4 3 3 b 3 + 6 5 4 3 3 4 b 4 +6b 5 + b 6 6 +6 5 b +5 4 b +0 3 b 3 +5 b 4 +6b 5 + b 6 74. Using the Binomil Theorem with k 7we hve ( + b) 7 7 +7 6 b + 7 6 5 b + 7 6 5 3 4 b 3 + 7 6 5 4 3 4 3 b 4 + 7 6 5 4 3 3 4 5 b 5 +7b 6 + b 7 7 +7 6 b + 5 b +35 4 b 3 +35 3 b 4 + b 5 +7b 6 + b 7 75. Using the Binomil Theorem with, b, k 4we hve ( ) 4 [ +( )] 4 ( ) 4 +4( ) 3 ( ) + 4 3 ( ) ( ) +4( )( ) 3 +( ) 4 8 4 6 +6 4 4 + 76. Using the Binomil Theorem with 3, b, k 5we hve (3 + ) 5 3 5 +5(3) 4 ( ) + 5 4 (3)3 ( ) + 5 4 3 3 (3) ( ) 3 +5(3)( ) 4 +( ) 5 43 + 405 +70 4 +90 6 +5 8 + 0 77. UsingEqution0, 3 3 64 8. 3 3 78. 3 3 54 54 3 7 3 7 3 3 79. UsingEqution0, 4 3 4 4 4 3 4 4 4 4 3 80. y 3 y (y)( 3 y) 4 y y 4 4 4 6. 8. UsingEqution0, 6 4 b 3 6 4 b 3 4 b 3/ 4 bb / 4 b b. 5 96 6 96 6 8. 5 5 3 3 5 3 5 83. UsingLws3ndofEponentsrespectively,3 0 9 8 3 0 (3 ) 8 3 0 3 8 3 0 + 6 3 6. 84. UsingLws3nd, 6 4 0 6 6 6 ( ) 0 ( 4 ) 6 6 0 4 60. 85. Using Lws 4,, nd of Eponents respectively, 9 () 4 9 ( 4 ) 4 69+4 6 9+4 3 6 0. 3 3 3 86. UsingLwsnd, n n + n +n + 3n + n n n 3n + (n ) n +3. 87. Using Lw of Eponents, 3 b 4 5 b 5 3 ( 5) b 4 5 b b. + y 88. ( + y) ( + y) + y + (y + ) ( + y) y y y 89. By definitions 3 nd 4 for eponents respectively, 3 / 3 / 3. 90. 96 /5 5 96 5 3 3 5 3 5 3 5 3 9. Using definition 4 for eponents, 5 /3 3 5 5 5. 9. 64 4/3 64 4/3 3 64 4 4 4 56

0 REVIEW OF ALGEBRA 3 3 93. ( y 4 ) 3/ 3/ ( ) 3/ (y 4 ) 3/ / y 4 3 (y ) 3 3 y 6 94. ( 5 y 3 z 0 ) 3/5 ( 5 ) 3/5 (y 3 ) 3/5 (z 0 ) 3/5 5/5 y 9/5 z 30/5 y 9/5 z 6 95. 5 y 6 y 6/5 by definition 4 for eponents. 96. ( 4 ) 3 ( /4 ) 3 3/4 97. 98. 99. t 5 8 5 4 5/8 3 4 t / st s /3 (t / ) 5 t 5/ t 5/ 3/4 (5/8) (3/4) /8 t / s / t / s /3 /8 /4 t (/) + (/) s (/) (/3) /4 (ts /6 ) /4 00. 0. 0. t /4 s ( /6) (/4) t/4 s /4 4 r n + 4 r 4 r n + r 4 r n + 4 r n (r n ) /4 r n/4 r n/ 3 3 +3 9 9 ( 9) +3 ( 9) ( +3) +3 + + ( ) + ( ) + + + 03. 8 4 8 4 +8 +8 3 64 ( 4)( +8) ( 4)( +4 +6) ( 4)( [Eqution 4 with, b 4] +4 +6 +8) +8 +h + h +h + h +h h +h ( h) 04. h h +h h h +h h +h h 05. 3 5 3 5 3+ 5 3+ 5 3+ 5 3+ 5 9 5 + y + y 06. y y + y y 07. +3 +4 +3 +4 +3 +4+ +3 +4+ +3 +4 +3 +4+ 3 +4 +3 +4+ 08. + + + + + + + ( ) + + 09. Flse. See Emple 4(b). + + 0. Flse. See the wrning fter Eqution 0.. True: 6 + 6. Flse: 3. Flse. 6 6 + 6 + 6 + y y + y + y 4. Flse. See the wrning on pge. y + y 6 + y Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 5. Flse. Using Lw 3 of Eponents, ( 3 ) 4 3 4 6 7. 6. True. 7. 5 3 8 8 8. π π becuse π > 0. 9. 5 5 5 5 5 5 becuse 5 5 < 0. 0. 3 3. If <, < 0,so ( ).. If >, > 0,so. + if + 0 3. + ( +) if +< 0 + if if < 4. if 0 ( ) if < 0 if if < 5. + +(since + 0 for ll ). 6. Determine when < 0 < > > if < or >.Thus, if < or > 7. +7> 3 > 4 >,so (, ). > 8. 4 3 6 3 3,so, 3. 9.,so [, ). 30. +5>5 3 8 >4 >,so,. 3. 0 < <0 >0,so (0, ]. 3. < 3 +4 6 3 < 3 < 4, so (, 4]. Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. 33. ( )( ) > 0. Cse : (both fctors re positive, so their product is positive) > 0 >,nd > 0 >,so (, ). Cse : (both fctors re negtive, so their product is positive) < 0 <,nd < 0 <,so (, ). Thus, the solution set is (, ) (, ). 34. < +8 8 < 0 ( 4)( +)< 0. Cse : >4 nd <, which is impossible. Cse : <4 nd >. Thus, the solution set is (, 4). 35. < 3 3 < 0 3 + 3 < 0. Cse : > 3 nd < 3, which is impossible. Cse : < 3 nd > 3. Thus, the solution set is 3, 3. Another method: < 3 < 3 3 << 3.

REVIEW OF ALGEBRA 36. 5 5 0 5 + 5 0. Cse : 5 nd 5,so 5,. Cse : 5 nd 5,so, 5. Thus, the solution set is, 5 5,. Another method: 5 5 5 or 5. 37. 3 0 ( ) 0. Since 0 for ll, the inequlity is stisfied when 0. Thus, the solution set is (, ]. 38. ( +)( )( +3)0,,or 3. Construct chrt: Intervl + +3 ( +)( )( +3) < 3 3 << + + << + + > + + + + Thus, ( +)( )( +3) 0 on [ 3, ] nd [, ), nd the solution set is [ 3, ] [, ). 39. 3 > 3 >0 > 0 ( )( +)> 0. Construct chrt: Intervl + ( )( +) < <<0 + + 0 << + + > + + + + Since 3 >when the lst column is positive, the solution set is (, 0) (, ). 40. 3 +3<4 3 4 +3<0 4 +3 < 0 ( )( 3) < 0. Thus, the solution set is (, 0) (, 3). Intervl 3 ( )( 3) <0 0 << + + <<3 + + >3 + + + + 4. / < 4. This is clerly true for <0. So suppose >0. then/ < 4 < 4 4 <.Thus,the solution set is (, 0) 4,. Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved.

REVIEW OF ALGEBRA 3 4. 3 < /. We solve the two inequlities seprtely nd tke the intersection of the solution sets. First, 3 < / is clerly true for >0. So suppose <0. Then 3 < / 3 > <,soforthis 3 inequlity, the solution set is, 3 (0, ). Now/ is clerly true if <0. So suppose >0. Then /, nd the solution set here is (, 0) [, ). Tking the intersection of the two solution sets gives the finl solution set:, 3 [, ). 43. C 5 9 (F 3) F 9 5 C +3.So50 F 95 50 9 5 C +3 95 8 9 5 C 63 0 C 35. So the intervl is [0, 35]. 44. Since 0 C 30 nd C 5 (F 3), wehve0 5 (F 3) 30 36 F 3 54 9 9 68 F 86. So the intervl is [68, 86]. 45. () Let T represent the temperture in degrees Celsius nd h the height in km. T 0when h 0nd T decreses by 0 Cforeverykm( C for ech 00-m rise). Thus, T 0 0h when 0 h. (b) From prt (), T 0 0h 0h 0 T h T/0. So0 h 5 0 T/0 5 T/0 3 0 T 30 0 T 30 30 T 0. Thus, the rnge of tempertures (in C) to be epected is [ 30, 0]. 46. Thebllwillbetlest3 ft bove the ground if h 3 8 + 6t 6t 3 6t 6t 96 0 6(t 3)(t +) 0. t 3nd t re endpoints of the intervl we re looking for, nd constructing tble gives t 3. Butt 0, so the bll will be t lest 3 ft bove the ground in the time intervl [0, 3]. 47. +3 + either +3 +or +3 ( +).Inthefirst cse,, nd in the second cse, +3 3 4 4. So the solutions re 4 nd. 3 3 48. 3 +5 either 3 +5or. Inthefirst cse, 3 4 4, nd in the second cse, 3 3 6. So the solutions re nd 4 3. 49. By Property 5 of bsolute vlues, < 3 3 < < 3, so ( 3, 3). 50. By Properties 4 nd 6 of bsolute vlues, 3 3 or 3, so (, 3] [3, ). 5. 4 < < 4 < 3 <<5,so (3, 5). 5. 6 < 0. 0. < 6 < 0. 5.9 <<6.,so (5.9, 6.). Stewrt: Clculus, Sith Edition. ISBN: 04950606. 008 Brooks/Cole. All rights reserved. 53. +5 +5 or +5 3 or 7,so (, 7] [ 3, ). 54. + 3 + 3 or + 3 or 4,so (, 4] [, ). 55. 3 0.4 0.4 3 0.4.6 3.4.3.7,so [.3,.7]. 56. 5 < 6 6 < 5 < 6 4 < 5 <8 4 5 << 8 5,so 4 5, 8 5. 57. (b c) bc b c bc b bc 58. + b<c < c b > c b 59. b (b) b b b + c bc + c (since <0) bc + c b 60. If 0 <<b,then < b nd b<b b [using Rule 3 of Inequlities]. So <b<b nd hence <b.