College algebra with intermediate algebra a blended course 1st edition beecher solutions manual down

Solution Manual for College Algebra with Intermediate Algebra A

Blended Course 1st Edition by Beecher Penna Johnson Bittinger

ISBN 0134555260

9780134555263

Full download link at:

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Chapter 4

Polynomials and Polynomial Functions

Degrees of terms: 2, 5, 7, 0

Degreeofpolynomial: 7 Leadingterm: a4b3 Leadingcoefficient: 1

Degrees of terms: 6, 8,4,2, 0

Degreeofpolynomial:8 Leading term:5p4w4

There are 140 golf balls in the third stack.

26. a) I(22)= 00560(22)4 +79980(22)3 4361840(22)2 +11, 6278376(22) 90, 625011 ≈ $26, 119

b) I(40) ≈ $45, 000

28. C(x)=5000+0.4x2

a) C(50) = 5000+04(50)2 = $6000

b) C(95) = 5000+04(95)2 = $8610 30. P (x) = 280x 07x2 (8000+05x2) P (x) = 280x 1.2x2 8000, or 12x2 +280x 8000 32. M = 162 84 64+1 = 15

92.

94. The area of the base is x x, or x2 .

The area of each side is x · (x 2).

The total area of all four sides is 4x(x 2).

The area of the top is x x, or x2 .

Thesurfaceareaofthisboxcanbeexpressedasapolynomial.

A = P + Pi + 4

(a b)(a2 + ab + b2) = (a b)(a2)+(a b)(ab)+(a b)(b2) = a3 a2b + a2b ab2 + ab2 b3

84. (3a 2b + c)(3a 2b c)

= (3a 2b)2 c2

= 9a2 12ab +4b2 c2

86. ( )=4 +2 2

90. f(x)= 4x 2x2

f(t 1) = 4(t 1) 2(t 1)2

= 4t 4 2(t2 2t +1)

= 4t 4 2t2 +4t 2

f x x x 2 = 2t2 +8t 6

f(t 1) = 4(t 1)+2(t 1)

= 4t 4+2(t2 2t +1)

= 4t 4+2t2 4t +2

= 2t2 2

f(p +1) = 4(p +1)+2(p +1)2

= 4p +4+2(p2 +2p +1)

= 4p +4+2p2 +4p +2

= 2p2 +8p +6

f(a + h) f(a)

= [4(a + h)+2(a + h)2] (4a +2a2)

= 4a +4h +2(a2 +2ah + h2) 4a 2a2

= 4a +4h +2a2 +4ah +2h2 4a 2a2

= 2h2 +4h +4ah 2

f(t 2)+ c = 4(t 2)+2(t 2) + c

= 4t 8+2(t2 4t +4)+ c

= 4t 8+2t2 8t +8+ c = 2t2 4t + c

f(a)+5 = 4a +2a2 +5

f(p +1)= 4(p +1) 2(p +1)2

= 4p +4 2(p2 +2p +1)

= 4p +4 2p2 4p 2

= 2p2 +2

f(a + h) f(a)

= [4(a + h) 2(a + h)2] [4a 2a2]

= 4a +4h 2(a2 +2ah + h2) 4a +2a2

= 4a +4h 2a2 4ah 2h2 4a +2a2

= 2h2 +4h 4ah

f(t 2)+ c = 4(t 2) 2(t 2)2 + c

= 4t 8 2(t2 4t +4)+ c

= 4t 8 2t2 +8t 8+ c

= 2t2 +12t 16+ c

f(a)+5 = 4a 2a2 +5 92. f(x)=2 4x 3x2

f(t 1) = 2 4(t 1) 3(t 1)2

2 4t +4 3(t2 2t +1)

f x x x = 3t2 +2t +3

f(t 1) = 3(t 1)2 4(t 1)+7

3(t2 2t +1) 4t +4+7

3t2 6t +3 4t +4+7

f(p +1)= 2 4(p +1) 3(p +1)2

= 2 4p 4 3(p2 +2p +1)

= 2 4p 4 3p2 6p 3 = 3p2 10p 5

f(p +1) = 3(p +1) 4(p +1)+7

3(p2 +2p +1) 4p 4+7

3p2 +6p +3 4p 4+7

3p2 +2p +6

f(a + h) f(a)

[3(a + h)2 4(a + h)+7] [3a2 4a +7] = 3(a2 +2ah + h2) 4a 4h +7 3a2 +4a 7 = 3a2 +6ah +3h2 4a 4h +7 3a2 +4a 7

= 3h2 +6ah 4h

f(t 2)+ c

f(a + h) f(a)

= [2 4(a + h) 3(a + h)2] [2 4a 3a2]

= 2 4a 4h 3(a2 +2ah + h2) 2+4a +3a2

= 2 4a 4h 3a2 6ah 3h2 2+4a +3a2

= 3h2 4h 6ah

f(t 2)+ c = 2 4(t 2) 3(t 2)2 + c

= 2 4t +8 3(t2 4t +4)+ c

= 2 4t +8 3t2 +12t 12+ c

= 3t2 +8t 2+ c 2

=3(t 2)2 4(t 2)+7+ c f(a)+5 = 2 4a 3a +5

94. Let r =thespeedof theplanein stillair.

Solve: d = (r 20)5, d = (r +20)4

Wefindthat r =180mph.

Multiply (1) by 5 and add.

5x 20y = 65

5x 7y = 16 27y = 81

Substitute3for y in(1)andsolvefor x

100. One method is as follows. For each equation, let y1 represent the left-hand side and y2 represent the right-hand side,andlet y3 = y2 y1.Thenuseagraphingcalculatorto viewthegraphof y3 and/oratableofvaluesfor y3.If y3 = 0, the equation is correct. If y3 = 0, the equation is not correct.

Not correct

Exercise Set 4.3

RC2. In the expression x(x 2), x and x 2 are factors RC4. A polynomial that cannot be factored is said to be prime.

RC6. When we factor by grouping, we look for a common binomial factor.

(y + z)+ w(y + z) = (y + z)(x + w)

46. y3 y2 +3y 3 = y2(y 1)+3(y 1)

All real numbers, or (−∞ , ∞)

64. a3b7 = (ab4 c2)

Firstfindtheterminthesecondblank. Thisistheexpression that is multiplied with ab4 to obtain a3b7 , or a2b3 .

Then the termin the firstblank isthe productof a2b3 and c2 , or a2b3c2

We have a3b7 a

66. 3a2 +6a +30+

b

(ab4 c2).

Exercise Set 4.4

RC2. To factor x2 11x 12, we look for a factorizationof 12 in which the negative factor has the greater absolute value (because the coefficient of the middle term is negative). The answer is (d).

RC4. To factor x2 12x +20, we look for a factorizationof 20 in which both factors are negative (because the coefficientofthemiddletermisnegative).Theanswer is (b).

12. 10y + y2 +24= y2 +10y +24=(y +4)(y +6)

14. 2 2 1 1 1

p + 5p + 25 = p + 5 p + 5

16. y2 14y +45 = (y 5)(y 9)

18. x + x2 90 = x2 + x 90 =(x +10)(x 9)

20. y2 +8y +7=(y +7)(y +1)

22. 32+4y y2 = y2 +4y +32 =

1(y2 4y 32)= 1(y 8)(y +4), or ( y +8)(y +4), or (y 8)( y 4)

24. 56x + x2 x3 = x3 + x2 +56x = x(x2 x 56) = x(x 8)(x +7), or x( x +8)(x +7), or x(x 8)( x 7)

26. y4 +5y2 84= (y2 +12)(y2 7)

28. x2 +12x +13 is not factorable into binomials with integer coefficients.

30. p2 5pq 24q2 = (p 8q)(p +3q)

32. 3x2 21x 90=3(x2 7x 30)=3(x 10)(x +3) 34. 24 a2 10a =

56. x2 + qx 32

Allsuch q arethesumsof thefactors of 32.

x3 +6)

40. x8 7x4 +10 = (x4 2)(x4 5) 42. a2 +14a +049=(a +07)(a +07)

44. 8 7t15 t30 = t30 7t15 +8 = 1(t30 +7t15 8) = 1(t15 +8)(t15 1), or ( t15 8)(t15 1), or (t15 +8)( t15 +1)

46. Let x and y representthenumberofweekdaysandweekend days Takako worked, respectively.

Solve: x + y =17

50x +60y = 940

The solution is (8, 9), so Takako worked 8 weekdays.

48. The graph isnot that of a function, because a vertical line can cross the graph at more than one point.

50. The graph is that of a function, because no vertical line

can cross the graph at more than one point.

q canbe31, 31,14, 14,4,or 4.

58. Left to the student

Chapter 4 Mid-Chapter Review

1. True; 5x +2x2 4x3 = x(5+2x 4x2).

2. The statement is false because one of the variables has a negative exponent. See page 218 in the text.

3. True; see page 219 in the text.

4. False; (

Degreeofeachterm: 7,4,1( a = a1),0(8=8a0)

Degree of the polynomial: 7

Leading term: a7

Leading coefficient: 1 ( a7 = 1 a7)

Constant term:8

7. 3x4 +2x3w5 12x2w +4x2 1

Terms: 3x4 , 2x3w5 , 12x2w, 4x2 , 1

Degree ofeach term: 4, 8(3+5=8),3(2+1=3), 2, 0

Degree of the polynomial: 8

Leadingterm:2x3w5

Leadingcoefficient: 2 Constant term: 1

and 3.

2 4b +21

1(b2 +4b 21) = 1(b +7)(b 3) = ( b 7)(b 3) Multiplying b 7 by 1 =(b +7)( b +3) Multiplying b +3 by 1

26. (9x 4)(9x +4)= (9x)

27. 5h2 +7h = h 5h + h 7= h(5h +7)

16

Thereisnopairoffactorsof6whosesumis8. This trinomial is not factorable into binomials.

36. Oneexplanationisasfollows.Theexpression (a b) is the opposite of a b. Since (a b)+(b a) = 0, then (a b)= b a

37. No;ifthecoefficientsofat least one pair of like terms are opposites, then the sum is a monomial.For example, (2x +3)+( 2x +1) =4, a monomial.

38. No;considerthepolynomial3x11 +5x7.Allofthecoefficientsandexponentsareprimenumbers,yetthepolynomial can be factored so it is not prime.

39. When coefficients and/or exponents are large, a polynomial is more easily evaluated after it has been factored.

40. a) The middle term, 2 a 3, ismissing from the right-hand side.

(a +3)2 = a2 +6a +9

b) The middle term, 2ab, is missing from the right-hand side and the sign preceding b2 is incorrect.

(a b)(a b)= a2 2ab + b2

c) The product of the outside terms and the product oftheinsidetermsaremissingfromtheright-hand side.

(x +3)(x 4)= x2 x 12

d) Thereshouldbeaminussignbetweenthetermsof the product.

p +7)(p 7) = p2 49

e) The middle term, 2 t 3,ismissingfromtherighthand side and the sign preceding 9 is incorrect.

41. Answermayvary.Forthepolynomial4a3 12a,anincor-rect factorization is 4a(a 3). Evaluating both the poly- nomial andthefactorizationfor a =0,weget0ineachcase.Thus, the evaluation does not catch the mistake.

Exercise Set 4.5

RC2. The product of the last terms mustbe1.

RC4. Bothlasttermsmustbenegative(becausethecoefficient of the middle term is negative).

RC6. Findtwointegerswhoseproductis20andwhosesum is 21. The integers are 20 and 1.

RC8. Factor by grouping.

2. 8x2 6x 9=(4x +3)(2x 3)

16(3)2 +96(3)+880=1024 ft

h(8)= 16(8)2 +96(8)+880=624ft

h(10)= 16(10)2 +96(10)+880=240 ft

b) h(t)= 16t2 +96t +880

y

Multiply(1)by2andadditto(2).

4x + 2y + 4z = 10

4 2 3 = 5 = 5(3a 1)(2a 5)

16. 18y2 +51y +15 = 3(6y2 +17y +5) =3(3y +1)(2y +5)

18. 15x3 19x2 10x = x(5x +2)(3x 5)

20. 70x4 68x3 +16x2 = 2x2(5x 2)(7x 4)

22. 6a2 7a 10 = (6a +5)(a 2)

24. 6y2 y 2 = (3y 2)(2y +1)

x y z

8x + z = 15 (4)

Add (1) and (3).

2x + y + 2z = 5

8x y + z = 5

6x + 3z = 0 (5)

Multiply (4) by 3 and add it to (5).

24x 3z = 45

6x + 3z = 0

26. 8 6a 9a2 = 9a2 6a +8 = 1(9a2 +6a 8) = 1(3a 2)(3a +4), or 30x = 45 3 ( 3a +2)(3a +4), or (3a 2)( 3a 4)

28. 12a2 +7a 1= 1(4a 1)(3a 1), or

x = 2

Substitute 3 for x in(4) andsolvefor z 2

Substitute 2 for x and 4 for y in (2) and solve for z 2

+12+ z = 17 z +16=17 z = 1

The solution is (2, 4, 1).

58. Write each equation in slope-intercept form. 7 y = 7x 3

= 7x 9

Since the lines have the same slope and different yintercepts, they are parallel.

60. Write each equation in slope-intercept form:

Sincetheproductoftheslopesis 1,thelinesareperpendicular.

Exercise Set 4.6

x2 25) = x(3x +5)(3x 5)

46. 2a9 32a =2a(a8 16)=2a(a4 +4)(a4 4)=

a(a4 +4)(a2 +2)(a2 2) 48. 16x6 121x2y4 = x2(16x4 121y4)=

2(4x2 +11y2)(4x2 11y2)

50. 1 2 = 1 + 1 100 10 10

ab3 +125a = a(b3 +125) = a(b +5)(b2 5b +25)

92. 2y3z4 54z7 = 2z4(y3 27z3) = 2z4(y 3z)(y2 +3yz +9z2)

94. 125c6 8d6 = (5c2 2d2)(25c4 +10c2d2 +4d4)

96. t6 +1= (t2 +1)(t4 t2 +1)

98. p6 q6 = (p3 + q3)(

Substitute 2 for m, 1 for x, and 7 for y in y = mx + b

Line perpendicular to x 4y =10:

The slope is 4. Substitute to find b.

y = mx + b

0 = 4 · 6+ b 24 = b

Theequation is y = 4x +24.

118. If P (x)= x4 , then P (a + h) P (a) = (a + h)4 a4 = [(a + h)2 + a2][(a + h)2 a2] = [(a + h)2 + a2][(a + h)+ a][(a + h) a] = (a2 +2ah + h2 + a2)(2a + h)(h)

= h(2a + h)(2a2 +2ah + h2)

120. The model shows a cube with volume a3 from which a portionwhosevolumeis b3 hasbeenremoved. Thisleaves a remaining volume which can be expressed as a2(a b)+ ab(a b

(a b), or (a b)(a2 +ab+b2). Thus, a3 b3 =

Exercise Set 4.7

RC2. If there are two terms, determine whether the binomialisadifferenceofsquares,asumofcubes, oradifferenceof cubes.

RC4. Iftherearefourterms,tryfactoringbygrouping RC6. Always check by multiplying.

= (a + b)2(a b)

b2)

28. 37x2 + x4 +36 = x4 37x2 +36

= (x2 36)(x2 1)

= (x +6)(x 6)(x +1)(x 1)

30. xw yw + xz yz

= w(x y)+ z(x y)

= (x y)(w + z)

32. 9c2 +12cd 5d2 =(3c d)(3c +5d)

34. 9m2 +3m3 +8m +24

= 3m3 +9m2 +8m +24

= 3m2(m +3)+8(m +3)

= (m +3)(3m2 +8)

36. 3x3 +6x2 27x 54

= 3(x3 +2x2 9x 18)

= 3[x2(x +2) 9(x +2)]

= 3(x +2)(x2 9)

6

50. 23xy +20x2y2 +6

= 20x2y2 23xy +6

= (4xy 3)(5xy 2)

52. 64p4 p

= p(64p3 1)

= p(4p 1)(16p2 +4p +1)

54. m2 n2 8n 16

= m2 (n2 +8n +16)

= m2 (n +4)2

= [m +(n +4)][m (n +4)]

= (m + n +4)(m n 4)

56. y2 14y +49 z2 = (y 7)2 z2

= (y 7+ z)(y 7 z)

58. Let x and y represent the length of a side of the pentagon and the octagon, respectively.

Solve:5x =8y,

Exercise Set 4.8

RC2. False,seeExample4onpage269inthe text. RC4. False; seepage 274in thetext.

46. x2 +14x +50= 5

x2 +14x +45= 0

(x +9)(x +5)= 0

x +9 = 0 or x +5 = 0

x = 9 or x = 5

The values of x for which f(x)=5 and 9 and 5.

48. 2x2 15x = 7 2x2 15x +7 = 0

(2x 1)(x 7) = 0

2x 1 = 0 or x 7 = 0

x = 1 or x = 7

x = 2 or x = 7 1

Thevaluesof x forwhich g(x)= 7are and7.

50. 4x x2 = 32 0 = x2 4x 32

0 = (x 8)(x +4)

x 8 = 0 or x +4 = 0

x = 8 or x = 4

The values of x for which h(x)= 32 are 8 and 4.

52. f(x)= 2 x2 7x +6

To find the excluded values, solve:

x2 7x +6 = 0

0 = 3y(y2 49) 0 = 3y(y +7)(y 7) (x 6)(x 1) = 0

(x2 4)(x2 9) = 0(x

3y = 0 or y +7 = 0 or y 7 = 0

y = 0 or y = 7 or y = 7

40. x4 13x2 +36 = 0

+2)(x 2)(x +3)(x 3) = 0

x+2 = 0 or x 2= 0 or x+3 = 0 or x 3= 0

x = 2 or x = 2 or x = 3 or x = 3

To

Thedomainof f is 5

x|x isareal number and

68. Let x =thelengthofasideoftheoriginalsquare.

Solve (x +4)2 =49.

x = 11 or x =3

Length cannot be negative.

The length is 3 m.

70. Let x, x +2, and x +4 represent the integers.

Solve x2 +(x +4)2 =136.

x = 10 or x =6

Theintegersare 10, 8,and 6or6,8,and10.

72. Let w =thewidth. Then w +3=thelength.

Solve w(w +3)=108.

w = 12 or w =9

The width cannot be negative.

The width is 9 m, and the length is 9+3, or 12 m.

74. Let x =thewidthoftheframe.

x =2 or x = 15

The width of the frame cannot be 15 cm.

The width of the frame is 2 cm.

76. Let w = the width. Then w +50= the length.

Solve w2 +(w +50)2 =2502

w = 200 or w =150

The width cannot be negative.

The width is 150 ft, and the length is 150+50, or 200 ft.

78. Solve 52 + x2 =(x +1)2

x = 12, so the length of the longer leg is 12 cm and the length of the hypotenuse is 12+1, or13, cm.

80. Let x =thelengthofoneleg.Then x +2=thelengthofthe hypotenuse, and (x + 2) 9, or x 7 = the length of the other leg.

60. From the graph we see that the x-intercepts are ( 8, 0)and (5, 0).Thesolutionsoftheequationarethefirstcoordinates of the x-intercepts, 8 and5.

62. From the graph we see that the x-intercepts are (0, 0) and (4, 0).Thesolutionsoftheequationarethe first coordi-nates of the x-intercepts, 0 and 4.

64. Let w =thewidth.Then w +4=the length.

Solve (w +4)w =96

w = 12 or w =8

The width cannot be negative.

The width is 8 cm and the length is 8+4, or 12 cm.

66. Let b represent the length of the base.

Solve x2 +(x 7)2 =(x +2)2

x =3 or x = 15

Since37isnegative,3cannotbeasolution.If x =15,then x +2=15+2=17and x 7 =15 7 =8. Thelengthsof thelegsare8ftand15ft,andthelengthofthehypotenuse is 17 ft.

82. Solve h(t)=0, or 16t2 +80t +224=0.

t =7 or t = 2

Thetimecannotbenegative. Theflarereachedtheground after 7 sec.

84. |0 ( 1023)| = |1023| = 1023

Solve2b(b +9)=56.

b = 16or b =7

The lengthof the base cannot be negative. Thus, the length of the base is 7 m and the height is 7+9, or 16 m.

The equation is y = 26 x + 934

31 31

92. Thefirstcoordinatesofthe x-interceptsare0and3. Then the solution set of x4 3x3 =0 is {0, 3}.

The graph lies on or below the x-axis for

x|0 ≤ x ≤ 3}, or [0, 3]. This is the solution set of x4 3x3 ≤ 0.

The graph lies above the x-axis for {x|x < 0 or x > 3},or(−∞ , 0) ∪ (3, ∞).Thisisthe solution set of x4 3x3 > 0.

94. a) (8x +11)(12x2 5x 2) = 0

(8x +11)(3x 2)(4x +1) = 0

8x+11 = 0 or 3x 2 = 0 or 4x+1=0

8x = 11 or 3x = 2 or 4x = 1 11 2 1 x = or x = or x =

Chapter 4 Vocabulary Reinforcement

1. Whenthetermsofa polynomial are written such thatthe exponents increase from left to right, we say that the polynomial is written in ascending order.

2. To factor a polynomial is to express it as a product.

3. A factor of a polynomial P is a polynomial that can be used to express P as a product.

4. Afactorizationofapolynomialisanexpressionthatnames that polynomial as a product.

5. Whenfactoringapolynomialwithfourterms,tryfactoring by grouping

6. A trinomial square is the square of a binomial. 8 11

Thesolutionsare 8 ,

3 4 2 1 3, and 4

b)(3x2 7x 20)(x 5) = 0 (3x +5)(x 4)(x 5)= 0

3x +5 = 0 or x 4 = 0 or x 5 = 0 3x = 5 or x = 4 or x = 5 5 x = 3 or x = 4 or x = 5

7. Theprinciple ofzeroproductsstatesthatif ab =0, then a =0 or b =0.

8. The factorization of a difference of squares is the product of the sum and difference of two terms.

Thesolutionsare 5 , 4, and 5. 3

Chapter 4 Concept Reinforcement

1. False; if ab =0 then a =0 or b =0.

2. True; 27 t3 =33 t3

3. False;the expression is not a binomial because y has anegative exponent.

= 3x2 +5x +4

2

f(a+h) f(a)=3(a + h)2 (a + h)+2 (3a2 a +

x = 3 or x = 2

The solutions are 7 and 2. 3

Chapter 4 Review Exercises

1. 3x6y 7x8y3 +2x3 3x2

a) The degrees of the terms are 6+1, or 7; 8+3, or 11; 3; and 2. The degree of the polynomial is 11.

b) The leading term is the term of highest degree, 7x8y3 . The leading coefficient is 7. c) 3x

8)

8. 3x2 +19x 72

There is no common factor (other than 1 or 1).The factorization will be of the form

x+ )(x+ )

Look for a pair of factors of middle term. We have 72 that yield the desired

9. 10x2 33x 7

Thereisnocommonfactor(otherthan1or 1).Multiply the leading coefficient and the constant: 10( 7) = 70.

Lookfora pairoffactors of 70 whose sumis 33. The

14) numbers we need are 35 and 2.Split the middle term and factor by grouping.

6. In 2010, t =2010 2008 =2. Locate 2 on the t-axis,and then move up to the graph. Now move horizontallyto the f (t)-axis. This locates a value of about 4.9, so we estimate that the number of children participating in football in 2010 was about 4.9 million.

11. (4a b +3c) (6a 7b 4c)

(4a b +3c)+( 6a +7b +4c)

2a +6b +7c 12. (9p2 4p +4) ( 7p2 +4p +4)

16p2 8p 13. (6x2 4xy + y2) (2x2 +3xy 2y2)

x

4xy + y2)+( 2x2 3xy +2y2)

= 4x2 7xy +3y

16. (4ab +3c)(2ab c)

f

(a 1) = (a 1)2 2(a 1) 7

a2 4a 4 f(a + h) f(a)

a2 +2ah + h2 2a 2h

=

x8 x6 +5x2 3

(5x +2)2 29. 4y2 16

4(y2 4)

23. f(x)= x2 2x 7 x x x x 32. 27x3 8 = (3x)3 23 Differenceofcubes = (3x 2)(9x2 +6x +4) 33. 0064b3 0125c3 = (04b)3 (05c)3 Differenceofcubes 20x4 28x3 + 12x2 = (04 05 )(016 2 +02 +025 2)

ofsquares = 4(y +2)(y 2) 30. ax +2bx ay 2by

35. 2z8 16z6

= 2z6(z2 8)

36. 54x6y 2y

= 2y(27z6 1) Differenceofcubes

= 2y(3x2 1)(9x4 +3x2 +1)

37. 1+ a3 Sum of cubes

= (1+ a)(1 a + a2)

38. 36x2 120x +100

= 4(9x2 30x +25) Trinomialsquare

= 4(3x 5)2

39. 6t2 +17pt +5p2

= (3t + p)(2t +5p) FOIL or ac-method

40. x3 +2x2 9x 18

= x2(x +2) 9(x +2) = (x +2)(x2 9)

Thesolutionsare 3 2

44. 8y2 =14y

8y2 14y = 0

2y(4y 7) = 0

3 and 2 3

2y = 0 or 4y 7 = 0

y = 0 or 4y = 7 7 y = 0 or y = 4

The solutions are 0 and 7 4

46. We set f(x) equal to 4.

x2 7x 40 = 4

x2 7x 44 = 0

(

x +4)(x 11) = 0

x +4 = 0 or x 11= 0

x = 4 or x = 11

The values of x for which f(x)=4 are 4 and 11.

47. ( )= x 3

f x

3x2 +19x 14

f (x) cannot be calculated for any x-value for which the denominator, 3x2 + 19x 14, is 0. To find the excluded values, we solve:

+19x 14

.

48. Familiarize Usingthelabelsonthedrawinginthetext, we let w =the widthof thephotographand w +3 =the length, in inches. Then the dimensions with the border added are w +2+2 and w +3+2+ 2, or w +4 and w +7.

Translate Weusetheformulafortheareaofarectangle, A = lw.

(w +7)(w +4)=108

Solve. Wesolve theequation.

(w +7)(w +4)=108

w2 +11w +28=108

w2 +11w 80 = 0

(w +16)(w 5) = 0

w +16 = 0 or w 5 = 0

w = 16 or w = 5

Check The width cannot be negative, so we check only 5. If the width of the photograph is 5 in., then the length

45. r2 = 16

r2 16 = 0

(r +4)(r 4) = 0

r +4 = 0 or r 4 = 0

r = 4 or r = 4

Thesolutionsare 4and4.

is 5+3, or 8 in. With the border added, the dimensionsare 5+2+2 and 8+2+2, or 9 in. and 12 in. The area is 9 · 12, or 108 in2 . The answer checks.

State.The length of the photograph is8 in., and thewidth is 5 in.

49. Familiarize Let x, x+2,and x+4representthe integers.

Translate The sum of the squares of the integers is 83,so we have x2 +(x +2)2 +(x +4)2 =83.

Solve Wesolve the equation.

55. [a (b 1)][(b 1)2 + a(b 1)+ a2]=

x = 0 or 8x +1 = 0 or 8x 1 = 0 x = 0 or 8x = 1 or 8x = 1

or 7.

Check. 7, 5,and 3areconsecutive oddintegersand x = 0 or x = 8 or x =

( 7)2 +( 5)2 +( 3)2 = 49+25+9 = 83. Also,

9+25+49=83. Bothanswerscheck.

State Theintegersare 7, 5,and 3or3,5,and7.

50. Familiarize Let s = the length of a side of the square. Then the area is s · s, or s2

Translate

The solutions are 0, 8, and 8

1 1 and 7 are consecutive odd integers and 3

57. Asumoftwosquarescanbefactoredwhenthereisacommonfactorthatisaperfectsquare. Forexample,consider

six times

Area is 7 more thelength of a side.

s2 = 7 + 6 s

Solve. Wesolve the equation.

s2 = 7+6s

s2 6s 7 = 0 (s 7)(s +1) = 0

s 7 = 0 or s +1 = 0

s = 7 or s = 1

Check Thelengthofasidecannotbenegative,so wecheck only7.Theareais72,or49,and7morethansixtimes 7 is 7+6 · 7, or 7+42, or 49. The answer checks.

State The length of a side of the square is 7.

51. t3 64 = t3 43 =(t 4)(t2 +4t +16)

ThecorrectchoiceisA.

52. hm +5hn gm 5gn = h(m +5n) g(m +5n)

= (m +5n)(h g)

ThecorrectchoiceisC.

53. 128x6 2y6

= 2(64x6 y6) Differenceofsquares

= 2(8x3 + y3)(8x3 y3) Sumofcubesand differenceofcubes

= 2(2x+ y)(4x2 2xy + y2)(2x y)(4x2+2xy + y2)

54. (x +1)3 (x 1)3 Differenceofcubes

= [(x + 1) (x 1)][(x + 1)2 + (x + 1)(x 1) + (x 1)2]

58. See the procedure on page 359 of the text.

59. Add the opposite of the polynomial being subtracted.

60. To solve P (x) = 0, find the first coordinate(s) of the xintercept(s) of y = P (x).

Tosolve P (x)=4,findthefirstcoordinate(s)ofthepoints of intersection of the graphs of y1 = P (x) and y2 = 4.

61. Tousefactoring,write x3 8 =(x 2)(x2 +2x +4)and(x 2)3 =(x 2)(x 2)(x 2).Since(x 2)(x2 +2x+4)=(x 2)(x 2)(x 2),then x3 8=(x 2)3.Tousegraphing,enter y1 = x3 8 and y2 = (x 2)3, and show that the graphs are different.

62. Both are correct. The factorizations are equivalent: (a b)(x y)

= 1(b a)( 1)(y x)

= ( 1)( 1)(b a)(y x)

= (b a)(y x)

63. x = 5 or x = 3

x 5 = 0 or x +3=0

(x 5)(x +3)= 0

x2 2x 15 = 0

There cannot be more than two solutions of a quadratic equation. This is because a quadratic equation is factorable into at most two different linear factors. Each of these has one solution when set equal to zero asrequired by the principle of zero products.

64. Thediscussioncouldincludethefollowingpoints:a)

We cannowsolvecertainpolynomial equations.

b) Whereasmostlinearequationshaveexactlyonesolution, non-linear polynomial equations can have more than one solution.

c) Weusedfactoringandtheprincipleofzeroproducts to solve polynomial equations.

Chapter 4 Test

3. In 2010, t =2010 1980 =30. Locate 30 on the t-axisand then move up to the graph. Now move horizontallytothe m(t)-axis. This locates a value of about 250, so we estimate that about 250 million tons of municipal solid waste was generated in 2010.

2) (x y)(xy) (x y)(y2)

= x · x2 y · x2 x · xy ( y)(xy) x · y2 ( y)(y2)

= x3 x2y x2y + xy2 xy2 + y3

= x3 2x2y + y3

14. 3m2 + 4m 2 m2 3m + 5 15m2 +20m 10

9m3 12m2 + 6m

3m4 4m3 + 2m2

3m4 13m3 + 5m2 + 26m 10

15. (4y 9)2

= (4y)2 2 · 4y · 9+92

(A B)2 = A2 2AB + B2

= 16y2 72y +81

16. (x 2y)(x +2y)

= x2 (2y)2 (A + B)(A B)= A2 B2

= x2 4y2

17. f(x)= x2 5x

f(a +10) = (a +10)2 5(a +10)

= a2 +

21. p2 12p 28

We look for a pair of factors of 28 whose sum is 12. The numbers we need are 14 and 2.

p2 12p 28=(p 14)(p +2)

22. 12m2 +20m +3

We will use the FOIL method.

1) Therearenocommonfactors(otherthan1or 1).

2) Factor the first term, 12m2 . The possibilities are (12m+ )(m+ ) and (6m+ )(2m+ ) and(4m+ )(3m+ ).

3) Factor the last term, 3. We need to consider only positive factors because both the middle term and the last term are positive. The factors are 3 and 1.

4) Lookforfactorsinsteps(2)and(3)suchthatthesum of the products is the middle term, 20m. Trial and error leads us to the correct factorization:

5) Factor by grouping.

Wemustincludethecommonfactortogetafactorization of the original trinomial.

24x2 46x +10=2(3x 5)(4x 1)

31. 16a7b +54ab7

= 2ab(8a6 +27b6)

= 2ab[(2a2)3 +(3b2)3]

= 2ab(2a2 +3b2)[(2a2)2 2a2 3b2 +(3b2)2] A3 + B3 = (A + B)(A2 AB + B2)

= 2ab(2a2 +3b2)(4a4 6a2b2 +9b4)

32. x2 18= 3x

x2 3x 18 = 0

(x 6)(x +3) = 0

x 6 = 0 or x +3 = 0

x = 6 or x = 3

Thesolutionsare6and 3.

33. 5y2 125 = 0

5(y2 25) = 0

5(y +5)(y 5) = 0

y +5 = 0 or y 5 = 0

y = 5 or y = 5

Thesolutionsare 5and5.

34. 2x2 +21 = 17x 2x2 +17x +21= 0 (2x +3)(x +7) = 0 2x +3 = 0 or x +7=0

3 x = 2 or x = 7

= (y +4+10t)(y +4 10t) 29. 20a

) = 5(2a + b)(2a b)

The solutions are 3 and 7.

35. We set f(x) equal to 11. 30. 24x2 46x +10 3

We will use the ac-method.

1) We factor out the common factor, 2. x 15x +11 = 11 3

5) = 0 2(12x2 23x +5)

2) Nowwefactorthetrinomial12

23 +5. Multiply

theleadingcoefficient,12, x x t,5. 12 5=60

3) Look for a factorization of 60 in which the sum of thefactorsisthecoefficientofthemiddleterm, 23. The factors we need are 20and 3.

4) Split the middle term as follows: 23x