esolutions manual - powered by cognero page 1 · &&6602'(/,1* ken throws the discus at a school...

Factor each polynomial, if possible. If the polynomial cannot be factored using integers, write prime .

1. 3x2 + 17x + 10

SOLUTION:

In this trinomial, a = 3, b = 17 and c = 10, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 3(10) or 30 and identify the factors with a sum of 17.

The correct factors are 2 and 15.

So, 3x2 + 17x + 10 = (3x + 2) (x +5).

Factors of 30 Sum 1, 30 31 2, 15 17 5, 6 30

2. 2x2 + 22x + 56

SOLUTION:

The GCF of the terms 2x2, 22x, and 56 is 2. Factor this first.

2x2

+22x + 56 = 2(x2 + 11x + 28) Distributive Property

In the trinomial, a = 1, b = 11 and c = 28, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 1(28) or 28 and identify the factors with a sum of 11.


So, 2x2 + 22x + 56= 2(x + 4)(x + 7).

Factors of 28 Sum 1, 28 29 2, 14 16 4, 7 11

3. 5x2 − 3x + 4

SOLUTION:

In this trinomial, a = 5, b = –3 and c = 4, so m + p is negative and mp is positive. Therefore, m and p must have different signs. List the factors of 5(4) or 20 and identify the factors with a sum of –3.

There are no factors of 20 with a sum of –3. This trinomial is prime.

Factors of 20 Sum 1, –20 –19 –1, 20 19 2, –10 –8 –2, 10 8 4, –5 –1 –4, 5 1

4. 3x2 − 11x − 20

SOLUTION:

In this trinomial, a = 3, b = –11 and c = –20, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–20) or –60 and identify the factors with a sum of –11.

The correct factors are 4 and –15 .

So, 3x2 − 11x − 20 = (3x +4)(x – 5).

Factors of –60 Sum 1, –60 –59 –1, 60 59 2, –30 –28 –2, 30 28 3, –20 –17 –3, 20 17 4, –15 –11 –4, 15 11 5, –12 –7 –5, 12 7 6, –10 –4 –6, 10 4

Solve each equation. Confirm your answers using a graphing calculator.

5. 2x2 + 9x + 9 = 0

SOLUTION: Factor the trinomial.

Solve the equation using the Zero Product Property. (2x + 3)(x + 3) = 0

The roots are and –3.

Confirm the roots using a graphing calculator. Let Y1 = 2x2 + 9x + 9 and Y2 = 0. Use the intersect option from the

CALC menu to find the points intersection.

Thus, the solutions are and –3.

6. 3x2 + 17x + 20 = 0



The roots are and –4 or about –1.667 and –4.

Confirm the roots using a graphing calculator. Let Y1 = 3x2 + 17x + 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points intersection.


7. 3x2 − 10x + 8 = 0


Solve the equation using the Zero Product Property.

The roots are and 2 or about 1.333 and 2.

Confirm the roots using a graphing calculator. Let Y1 = 3x2 –10x + 8 and Y2 = 0. Use the intersect option from the


Thus, the solutions are and 2.

8. 2x2 − 17x + 30 = 0


Solve the equation using the Zero Product Property. (2x – 5) (x – 6) = 0

The roots are and 6.

Confirm the roots using a graphing calculator. Let Y1 = 2x2 –17x + 30 and Y2 = 0. Use the intersect option from



9. CCSS MODELING Ken throws the discus at a school meet.

a. What is the initial height of the discus? b. After how many seconds does the discus hit the ground?

SOLUTION: a. The initial height of the discus is for the time t = 0.

The initial height of the discus is 5 feet. b. When the discus hits the ground, the height will be zero, so h = 0. Substitute 0 for h in the equation and factor the trinomial.

The solutions to the equation are and . Since time cannot be negative, the discus will hit the ground after or

2.5 seconds. Check by substituting 2.5 in for x in the original equation.

Therefore, the discus will hit the ground after 2.5 seconds.


10. 5x2 + 34x + 24

SOLUTION:



So, 5x2 + 34x + 24= (5x + 4)(x + 6).

Factors of 120 Sum 1, 120 121 2, 60 62 3, 40 43 4, 30 34 5, 24 29 6, 20 26 8, 15 23 10, 12 22

11. 2x2 + 19x + 24

SOLUTION:



So, 2x2 + 19x + 24= (2x + 3)(x + 8).

Factors of 48 Sum 1, 48 49 2, 24 26 3, 16 19 4, 12 16 6, 8 14

12. 4x2 + 22x + 10

SOLUTION:


4x2 + 22x + 10 = 2(2x

2 + 11x + 5) Distributive Property

In the trinomial, a = 2, b = 11 and c = 5, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the factors of 2(5) or 10 and identify the factors with a sum of 11.


So, 4x2 + 22x + 10= 2(2x + 1)(x + 5).

Factors of 10 Sum 1, 10 11 2, 5 7

13. 4x2 + 38x + 70

SOLUTION:


4x2

+38x + 70 = 2(2x2 + 19x + 35) Distributive Property



So, 4x2 + 38x + 70= 2(2x + 5)(x + 7).

Factors of 70 Sum 1, 70 71 2, 35 37 5, 14 19 7, 10 17

14. 2x2 − 3x − 9

SOLUTION:


The correct factors are 3 and –6.

So, 2x2 − 3x − 9= (2x + 3)(x − 3).

Factors of –18 Sum 1, –18 –17 –1, 18 17 2, –9 –7 –2, 9 7 3, –6 –3 –3, 6 3

15. 4x2 − 13x + 10

SOLUTION:

In this trinomial, a = 4, b = –13 and c = 10, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 4(10) or 40 and identify the factors with a sum of –13.

The correct factors are –5 and –8.

So, 4x2 − 13x + 10= (4x − 5)(x − 2).

Factors of 40 Sum –1, –40 –41 –2, –20 –22 –4, –10 –14 –5, –8 –13

16. 2x2 + 3x + 6

SOLUTION:

In this trinomial, a = 2, b = 3, and c = 6, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 2(6) or 12 and identify the factors with a sum of 3.

There are no factors of 12 with a sum of 3.

So, 2x2 + 3x + 6 is prime.

Factors of 12 Sum 1, 12 13 2, 6 8 3, 4 7

17. 5x2 + 3x + 4

SOLUTION:




Factors of 20 Sum 1, 20 21 2, 10 12 4, 5 9

18. 12x2 + 69x + 45

SOLUTION:


12x2




So, 12x2 + 69x + 45= 3(4x + 3)(x + 5).

Factors of 60 Sum 1, 60 61 2, 30 32 3, 20 23 4, 15 19 5, 12 17 6, 10 60

19. 4x2 − 5x + 7

SOLUTION:

In this trinomial, a = 4, b = –5, and c = 7, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 4(7) or 28 and identify the factors with a sum of –5.

There are no factors of 28 with a sum of –11.

So, 4x2 − 5x + 7 is prime.

Factors of 28 Sum –1,–28 –29 –2, –14 –16 –4, –7 –11

20. 5x2 + 23x + 24

SOLUTION:



So, 5x2 + 23x + 24= (5x + 8)(x + 3).

Factors of 120 Sum 1, 120 121 2, 60 62 3, 40 43 4, 30 34 5, 24 29 6, 20 26 8, 15 23 10, 12 22

21. 3x2 − 8x + 15

SOLUTION:

In this trinomial, a = 3, b = –8, and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the factors of 3(15) or 45 and identify the factors with a sum of –8.



Factors of 45 Sum –1,–45 –46 –3, –15 –18 –5, –9 –14

22. SHOT PUT An athlete throws a shot put with an initial upward velocity of 29 feet per second and from an initial height of 6 feet. a. Write an equation that models the height of the shot put in feet with respect to time in seconds. b. After how many seconds will the shot put hit the ground?

SOLUTION:

a. A model for the vertical motion of a projected object is given by h = −16t2 + vt + h0, where h is the height in feet,

t is the time in seconds, v is the initial velocity in feet per second, and h0 is the initial height in feet. The initial velocity

of the shot put, v, is 29 feet per second. The initial height of the shot put, h0, is 6 feet.

So, the equation h = −16t2 + 29t + 6 models the height of the shot put in feet with respect to the time in seconds. b. When the shot put hits the ground, the height will be zero, so h = 0.

The roots are and 2. Because time cannot be negative, the shot put will hit the ground after 2 seconds.


23. 2x2 + 9x − 18 = 0



Confirm the roots using a graphing calculator. Let Y1 = 2x2 +9x – 18 and Y2 = 0. Use the intersect option from the



24. 4x2 + 17x + 15 = 0

SOLUTION: Factor the trinomial then solve the equation using the Zero Product Property. .


Confirm the roots using a graphing calculator. Let Y1 = 4x2 +17x + 15 and Y2 = 0. Use the intersect option from



25. −3x2 + 26x = 16

SOLUTION: Factor the trinomial, then solve the equation using the Zero Product Property.


Confirm the roots using a graphing calculator. Let Y1 = -3x2 +26x and Y2 = 16. Use the intersect option from the



26. −2x2 + 13x = 15



Confirm the roots using a graphing calculator. Let Y1 = -2x2 +13x and Y2 = 15. Use the intersect option from the CALC menu to find the points intersection.


27. −3x2 + 5x = −2



Confirm the roots using a graphing calculator. Let Y1 = -3x2 +5x and Y2 = -2. Use the intersect option from the



28. −4x2 + 19x = −30






29. BASKETBALL When Jerald shoots a free throw, the ball is 6 feet from the floor and has an initial upward velocity of 20 feet per second. The hoop is 10 feet from the floor. a. Use the vertical motion model to determine an equation that models Jerald’s free throw. b. How long is the basketball in the air before it reaches the hoop? c. Raymond shoots a free throw that is 5 foot 9 inches from the floor with the same initial upward velocity. Will the ball be in the air more or less time? Explain.

SOLUTION:



of the basketball, v, is 20 feet per second. The initial height of the basketball, h0, is 6 feet. The height of the hoop, h,

is 10 feet.

So, the equation 10 = −16t2 + 20t + 6 models Jerald’s free throw.

b.

The roots are and 1. The basketball takes second to reach a height of 10 feet on its way up. The basketball

takes 1 second to reach a height of 10 feet on its way down. So, the basketball will be in the air 1 second before it reaches the hoop. c. The ball will be in the air less time because it starts closer to the ground so the shot will not have as far to fall.

30. DIVING Ben dives from a 36-foot platform. The equation h = −16t2 + 14t + 36 models the dive. How long will it take Ben to reach the water?

SOLUTION: When Ben reaches the pool, his height, h will be 0.

The roots are and 2. Because time cannot be negative, Ben will reach the water after 2 seconds.

31. NUMBER THEORY Six times the square of a number x plus 11 times the number equals 2. What are possible values of x?

SOLUTION:

Let x = a number. Then, 6x2 +11x = 2.

The possible values of x are –2 or .


32. −6x2 − 23x − 20

SOLUTION: First factor the number –1 out of each term in the polynomial.

Then factor the trinomial 6x2 + 23x + 20.



So, −6x2 − 23x − 20= −(2x + 5)(3x + 4).

Factors of 120 Sum 1, 120 121 2, 60 62 3, 40 43 4, 30 34 5, 24 29 6, 20 26 8, 15 23 10, 12 22

33. −4x2 − 15x − 14


−4x2 − 15x − 14 = –1(4x2 + 15x + 14)




So, −4x2 − 15x − 14 = −(x + 2)(4x + 7).

Factors of 56 Sum 1, 56 57 2, 28 30 4, 14 18 7, 8 115

34. −5x2 + 18x + 8


−5x2 + 18x + 8 = –1(5x2 – 18x – 8)

Then factor the trinomial 5x2 – 18x – 8.


The correct factors are –20 and 2 .

So, −5x2 + 18x + 8 = −(x – 4)(5x + 2).

Factors of –40 Sum –1, 40 39 1, –40 –39 –2, 20 18 2, –20 –18 –4, 10 6 4, –10 –6 –5, 8 3 5, –8 –3

35. −6x2 + 31x − 35


−6x2 + 31x − 35 = –1(6x2 – 31x + 35)

Then factor the trinomial 6x2 – 31x + 35.



So, −6x2 + 31x − 35 = −(2x − 7)(3x − 5).

Factors of 210 Sum –1, –210 –211 –2, –105 –107 –3, –70 –73 –5, –42 –47 –6, –35 –41 –7, –30 –37 –10, –21 –31 –14, –15 –29

36. −4x2 + 5x − 12

SOLUTION: First factor the number –1 out of each term in the polynomial. −4x2 + 5x − 12 = –1(4x2 – 5x + 12)



There are no factors of 48 with a sum of –5. So, −4x2 + 5x − 12 is prime.

Factors of 48 Sum –1, –48 –49 –2, –24 –26 –3, –16 –19 –4, –12 –16 –6, –8 –14

37. −12x2 + x + 20


−12x2 + x + 20 = –1(12x2 – x – 20)

Then factor the trinomial 12x2 – x – 20.


The correct factors are –16 and 15.

So, −12x2 + x + 20 = −(4x + 5)(3x – 4).

Factors of –240 Sum –1, 240 239 1, –240 –239 –2, 120 118 2, –120 –118 –3, 80 77 3, –80 –77 –4, 60 56 4, –60 –56 –5, 48 43 5, –48 –43 –6, 40 34 6, –40 –34 –8, 30 22 8, –30 –22 –10, 24 14 10, –24 –14 –12, 20 8 12, –20 –8 –15, 16 1 15, –16 –1

38. URBAN PLANNING The city has commissioned the building of a rectangular park. The area of the park can be

expressed as 660x2 + 524x + 85. Factor this expression to find binomials with integer coefficients that represent

possible dimensions of the park. If x = 8, what is a possible perimeter of the park?

SOLUTION:

So, (22x + 5)(30x + 17) represent the possible dimensions of the park.

Evaluate each dimension when x = 8 to find the possible length and width of the park when x = 8.

A possible perimeter of the park is 876 units.

39. MULTIPLE REPRESENTATIONS In this problem, you will explore factoring a special type of polynomial. a. GEOMETRIC Draw a square and label the sides a. Within this square, draw a smaller square that shares a vertex with the first square. Label the sides b. What are the areas of the two squares? b. GEOMETRIC Cut and remove the small square. What is the area of the remaining region? c. ANALYTICAL Draw a diagonal line between the inside corner and outside corner of the figure, and cut along this line to make two congruent pieces. Then rearrange the two pieces to form a rectangle. What are the dimensions? d. ANALYTICAL Write the area of the rectangle as the product of two binomials.

e. VERBAL Complete this statement: a2 − b2 = … Why is this statement true?

SOLUTION: a.

The area of a square is found using the formula A = s2. So, the area of the larger square is a

2 and the area of the

smaller square is b2.

b.

To find the area of the remaining region, subtract the area of the smaller square from the area of the larger square.

So, the area of the remaining region is a2 – b

2.

c.

The width is a – b and the length is a + b. d.

e . The figure with area a2 − b2 and the rectangle with area (a − b)(a + b) have the same area, so a2 − b2 = (a − b)(a + b).

40. CCSS CRITIQUE Zachary and Samantha are solving 6x2 − x = 12. Is either of them correct? Explain your

reasoning.

SOLUTION: Sample answer: By the Zero Product Property, the equation must be set equal to zero before it can be factored. Then each of the factors can be set equal to zero and solved for x.

So, Samantha is correct.

41. REASONING A square has an area of 9x2 + 30xy + 25y2 square inches. The dimensions are binomials with positive integer coefficients. What is the perimeter of the square? Explain.

SOLUTION:

The area of the square equals 9x2 + 30xy + 25y

2.

So, the length of each side of the square is (3x + 5y) inches.

The perimeter of the square is (12x + 20y) inches.

42. CHALLENGE Find all values of k so that 2x2 + kx + 12 can be factored as two binomials using integers.

SOLUTION: The values for k must be the sum of two negative factors or two positive factors of 2(12) or 24.

Factors of 24 Possible Values for k –1, –24 –25

1, 24 25 –2, –12 –14

2, 12 14 –3, –8 –11

3, 8 11 –4, –6 –10

4, 6 10

43. WRITING IN MATH What should you consider when solving a quadratic equation that models a real-world situation?

SOLUTION: Sample answer: A quadratic equation may have zero, one, or two solutions. If there are two solutions, you must consider the context of the situation to determine whether one or both solutions answer the given question. For example: Time cannot be negative. A solution stating that a person ran 60 miles per hour is unrealistic.

44. WRITING IN MATH Explain how to determine which values should be chosen for m and p when factoring a

polynomial of the form ax2 + bx + c.

SOLUTION:

Sample answer: If the trinomial factors, then ax2 + bx + c = ax2 + mx + px + c where m and n are two integers thathave a product equal to ac and a sum equal b. For example, given the trinomial 3x

2 + 14x + 15, a = 3, b = 14, and c = 15. Find two integers such that mp = 3(15) or 45 and m + p = 14. Since 5 × 9 = 45 and 5 + 9 = 14, the values of thevariables could be m = 5 and p = 9. Check to see if the trinomial factors using these values for m and p .

45. GRIDDED RESPONSE Savannah has two sisters. One sister is 8 years older than her, and the other sister is 2 years younger than her. The product of Savannah’s sisters’ ages is 56. How old is Savannah?

SOLUTION: Let s = Savannah’s age. Then, s + 8 = the age of her older sister and s – 2 = the age of her younger sister.

The roots are –12 and 6. Savannah’s age cannot be negative. So, Savannah is 6 years old.

46. What is the product of a3b

5 and a

5b

2?

A a8b

7

B a2b

3

C a8b

3

D a2b

7

SOLUTION:

Choice A is the correct answer.

47. What is the solution set of x2 + 2x − 24 = 0?

F {−4, 6} G {−3, 8} H {3, −8} J {4, −6}

SOLUTION:

The solutions are –6 and 4. Choice J is the correct answer.

48. Which is the solution set of x ≥ −2? A

B

C

D

SOLUTION: Because x is greater than or equal to –2, there should be a closed circle at –2, and the line should be shaded to the right of –2. Choice C is the correct answer.

Factor each polynomial.

49. x2 − 9x + 14

SOLUTION:

In this trinomial, b = –9 and c = 14, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 14, and look for the pair of factors and identify the factors with a sum of –9.


Factors of 14 Sum –1, –14 –15 –2, –7 –9

50. n2 − 8n + 15

SOLUTION:



Factors of 15 Sum –1, –15 –16 –3, –5 –8

51. x2 − 5x − 24

SOLUTION:

In this trinomial, b = –5 and c = –24, so m + p is negative and mp is negative. Therefore, m and p must be opposite signs. List the factors of –24, and look for the pair of factors and identify the factors with a sum of –5.


Factors of –24 Sum 1, –24 –23 –1, 24 23 2, –12 –10 –2, 12 10 3, –8 –5 –3, 8 5 4, –6 –2 –4, 6 2

52. z2 + 15z + 36

SOLUTION:

In this trinomial, b = 15 and c = 36, so m + p is negative and mp is positive. Therefore, m and p must both be positive.List the factors of 36, and look for the pair of factors and identify the factors with a sum of 15.


Factors of 36 Sum 1, 36 37 2, 18 20 3, 12 15 4, 9 13 6, 6 12

53. r2 + 3r − 40

SOLUTION:

In this trinomial, b = 3 and c = –40, so m + p is positive and mp is negative. Therefore, m and p must be opposite signs. List the factors of –40, and look for the pair of factors and identify the factors with a sum of 3.


Factors of –40 Sum 1, –40 –39 –1, 40 39 2, –20 –18 –2, 20 18 4, –10 –6 –4, 10 6 5, –8 –3 –5, 8 3

54. v2 + 16v + 63

SOLUTION:

In this trinomial, b = 16 and c = 63, so m + p is positive and mp is positive. Therefore, m and p must be opposite signs. List the factors of 63, and look for the pair of factors and identify the factors with a sum of 16.


Factors of 63 Sum 1, 63 64 3, 21 24 7, 9 16

Solve each equation. Check your solutions.

55. a(a − 9) = 0

SOLUTION:

The roots are 0 and 9. Check by substituting 0 and 9 in for a in the original equation.

and

The solutions are 0 and 9.

56. (2y + 6)(y − 1) = 0

SOLUTION:

The roots are –3 and 1. Check by substituting –3 and 1 in for y in the original equation.

and

The solutions are –3 and 1.

57. 10x2 − 20x = 0

SOLUTION:

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and


58. 8b2 − 12b = 0

SOLUTION:

The roots are 0 and . Check by substituting 0 and in for b in the original equation.

and

The solutions are 0 and .

59. 15a2 = 60a

SOLUTION:


and


60. 33x2 = −22x

SOLUTION:

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and


61. ART A painter has 32 units of yellow dye and 54 units of blue dye to make two shades of green. The units needed to make a gallon of light green and a gallon of dark green are shown. Make a graph showing the numbers of gallons of the two greens she can make, and list three possible solutions.

SOLUTION: Let x = the number of gallons of light green and let y = the number of gallons of dark green. Then, 4x + y ≤ 32 and x + 6y ≤ 54. Graph 4x + y = 32 and x + 6y = 54. Both lines are included in the solution of the system and should be solid.

Some possible solutions are 2 light, 8 dark; 6 light, 8 dark; 7 light, 4 dark. They fall in the shaded region which represents the solution set.

Solve each compound inequality. Then graph the solution set.

62. k + 2 > 12 and k + 2 ≤ 18

SOLUTION:

The solution set is {k |10 < k ≤ 16}. To graph the solution set, graph k ≤ 16 and graph k > 10. Then find the intersection.

and

63. d − 4 > 3 or d − 4 ≤ 1

SOLUTION:

The solution set is {d|d ≤ 5 or d > 7}. Notice that the graphs do not intersect. To graph the solution set, graph d ≤ 5 and graph d > 7. Then find the union.

or

64. 3 < 2x − 3 < 15

SOLUTION:

The solution set is {x|3 < x < 9}.

To graph the solution set, graph x < 9 and graph x > 3. Then find the intersection.

and

65. 3t − 7 ≥ 5 and 2t + 6 ≤ 12

SOLUTION:

The solution set is empty. A number cannot be both greater than or equal to 4 and less than or equal to 3.

and

66. h − 10 < −21 or h + 3 < 2

SOLUTION:

The solution set is {h|h < −1}. To graph the solution set, graph h < –1. All numbers less than –1 will also be less than –11.

or

67. 4 < 2y − 2 < 10

SOLUTION:

The solution set is {y |3 < y < 6}.

To graph the solution set, graph y < 6 and graph y > 3. Then find the intersection.

and

68. FINANCIAL LITERACY A home security company provides security systems for $5 per week, plus an installation fee. The total cost for installation and 12 weeks of service is $210. Write the point-slope form of an equation to find the total fee y for any number of weeks x. What is the installation fee?

SOLUTION: The slope of the equation of the line that represents the weekly cost is 5. The total cost for 12 weeks of service including installation is $210. So, the line passes through the point (12, 210).

The point-slope form of an equation to find the total fee y for any number of weeks x is y − 210 = 5(x − 12). Solve the equation for y to find the flat fee for installation.

So, the flat fee for installation is $150.

Find the principal square root of each number.69. 16

SOLUTION:

70. 36

SOLUTION:

71. 64

SOLUTION:

72. 81

SOLUTION:

73. 121

SOLUTION:

74. 100

SOLUTION:

eSolutions Manual - Powered by Cognero Page 1

8-7 Solving ax^2 + bx + c = 0


1. 3x2 + 17x + 10

SOLUTION:



So, 3x2 + 17x + 10 = (3x + 2) (x +5).

Factors of 30 Sum 1, 30 31 2, 15 17 5, 6 30

2. 2x2 + 22x + 56

SOLUTION:


2x2




So, 2x2 + 22x + 56= 2(x + 4)(x + 7).

Factors of 28 Sum 1, 28 29 2, 14 16 4, 7 11

3. 5x2 − 3x + 4

SOLUTION:



Factors of 20 Sum 1, –20 –19 –1, 20 19 2, –10 –8 –2, 10 8 4, –5 –1 –4, 5 1

4. 3x2 − 11x − 20

SOLUTION:



So, 3x2 − 11x − 20 = (3x +4)(x – 5).

Factors of –60 Sum 1, –60 –59 –1, 60 59 2, –30 –28 –2, 30 28 3, –20 –17 –3, 20 17 4, –15 –11 –4, 15 11 5, –12 –7 –5, 12 7 6, –10 –4 –6, 10 4


5. 2x2 + 9x + 9 = 0







6. 3x2 + 17x + 20 = 0







7. 3x2 − 10x + 8 = 0







8. 2x2 − 17x + 30 = 0















10. 5x2 + 34x + 24

SOLUTION:



So, 5x2 + 34x + 24= (5x + 4)(x + 6).

Factors of 120 Sum 1, 120 121 2, 60 62 3, 40 43 4, 30 34 5, 24 29 6, 20 26 8, 15 23 10, 12 22

11. 2x2 + 19x + 24

SOLUTION:



So, 2x2 + 19x + 24= (2x + 3)(x + 8).

Factors of 48 Sum 1, 48 49 2, 24 26 3, 16 19 4, 12 16 6, 8 14

12. 4x2 + 22x + 10

SOLUTION:


4x2 + 22x + 10 = 2(2x




So, 4x2 + 22x + 10= 2(2x + 1)(x + 5).

Factors of 10 Sum 1, 10 11 2, 5 7

13. 4x2 + 38x + 70

SOLUTION:


4x2




So, 4x2 + 38x + 70= 2(2x + 5)(x + 7).

Factors of 70 Sum 1, 70 71 2, 35 37 5, 14 19 7, 10 17

14. 2x2 − 3x − 9

SOLUTION:



So, 2x2 − 3x − 9= (2x + 3)(x − 3).

Factors of –18 Sum 1, –18 –17 –1, 18 17 2, –9 –7 –2, 9 7 3, –6 –3 –3, 6 3

15. 4x2 − 13x + 10

SOLUTION:



So, 4x2 − 13x + 10= (4x − 5)(x − 2).

Factors of 40 Sum –1, –40 –41 –2, –20 –22 –4, –10 –14 –5, –8 –13

16. 2x2 + 3x + 6

SOLUTION:




Factors of 12 Sum 1, 12 13 2, 6 8 3, 4 7

17. 5x2 + 3x + 4

SOLUTION:




Factors of 20 Sum 1, 20 21 2, 10 12 4, 5 9

18. 12x2 + 69x + 45

SOLUTION:


12x2




So, 12x2 + 69x + 45= 3(4x + 3)(x + 5).

Factors of 60 Sum 1, 60 61 2, 30 32 3, 20 23 4, 15 19 5, 12 17 6, 10 60

19. 4x2 − 5x + 7

SOLUTION:




Factors of 28 Sum –1,–28 –29 –2, –14 –16 –4, –7 –11

20. 5x2 + 23x + 24

SOLUTION:



So, 5x2 + 23x + 24= (5x + 8)(x + 3).

Factors of 120 Sum 1, 120 121 2, 60 62 3, 40 43 4, 30 34 5, 24 29 6, 20 26 8, 15 23 10, 12 22

21. 3x2 − 8x + 15

SOLUTION:




Factors of 45 Sum –1,–45 –46 –3, –15 –18 –5, –9 –14


SOLUTION:







23. 2x2 + 9x − 18 = 0






24. 4x2 + 17x + 15 = 0






25. −3x2 + 26x = 16






26. −2x2 + 13x = 15





27. −3x2 + 5x = −2






28. −4x2 + 19x = −30







SOLUTION:




is 10 feet.


b.







SOLUTION:




32. −6x2 − 23x − 20





So, −6x2 − 23x − 20= −(2x + 5)(3x + 4).

Factors of 120 Sum 1, 120 121 2, 60 62 3, 40 43 4, 30 34 5, 24 29 6, 20 26 8, 15 23 10, 12 22

33. −4x2 − 15x − 14


−4x2 − 15x − 14 = –1(4x2 + 15x + 14)




So, −4x2 − 15x − 14 = −(x + 2)(4x + 7).

Factors of 56 Sum 1, 56 57 2, 28 30 4, 14 18 7, 8 115

34. −5x2 + 18x + 8


−5x2 + 18x + 8 = –1(5x2 – 18x – 8)




So, −5x2 + 18x + 8 = −(x – 4)(5x + 2).

Factors of –40 Sum –1, 40 39 1, –40 –39 –2, 20 18 2, –20 –18 –4, 10 6 4, –10 –6 –5, 8 3 5, –8 –3

35. −6x2 + 31x − 35


−6x2 + 31x − 35 = –1(6x2 – 31x + 35)




So, −6x2 + 31x − 35 = −(2x − 7)(3x − 5).

Factors of 210 Sum –1, –210 –211 –2, –105 –107 –3, –70 –73 –5, –42 –47 –6, –35 –41 –7, –30 –37 –10, –21 –31 –14, –15 –29

36. −4x2 + 5x − 12





Factors of 48 Sum –1, –48 –49 –2, –24 –26 –3, –16 –19 –4, –12 –16 –6, –8 –14

37. −12x2 + x + 20


−12x2 + x + 20 = –1(12x2 – x – 20)




So, −12x2 + x + 20 = −(4x + 5)(3x – 4).

Factors of –240 Sum –1, 240 239 1, –240 –239 –2, 120 118 2, –120 –118 –3, 80 77 3, –80 –77 –4, 60 56 4, –60 –56 –5, 48 43 5, –48 –43 –6, 40 34 6, –40 –34 –8, 30 22 8, –30 –22 –10, 24 14 10, –24 –14 –12, 20 8 12, –20 –8 –15, 16 1 15, –16 –1




SOLUTION:






SOLUTION: a.




b.



2.

c.




reasoning.




SOLUTION:


2.




SOLUTION: The values for k must be the sum of two negative factors or two positive factors of 2(12) or 24.

Factors of 24 Possible Values for k –1, –24 –25

1, 24 25 –2, –12 –14

2, 12 14 –3, –8 –11

3, 8 11 –4, –6 –10

4, 6 10

43. WRITING IN MATH What should you consider when solving a quadratic equation that models a real-world situation?

SOLUTION: Sample answer: A quadratic equation may have zero, one, or two solutions. If there are two solutions, you must consider the context of the situation to determine whether one or both solutions answer the given question. For example: Time cannot be negative. A solution stating that a person ran 60 miles per hour is unrealistic.

44. WRITING IN MATH Explain how to determine which values should be chosen for m and p when factoring a

polynomial of the form ax2 + bx + c.

SOLUTION:

Sample answer: If the trinomial factors, then ax2 + bx + c = ax2 + mx + px + c where m and n are two integers thathave a product equal to ac and a sum equal b. For example, given the trinomial 3x

2 + 14x + 15, a = 3, b = 14, and c = 15. Find two integers such that mp = 3(15) or 45 and m + p = 14. Since 5 × 9 = 45 and 5 + 9 = 14, the values of thevariables could be m = 5 and p = 9. Check to see if the trinomial factors using these values for m and p .

45. GRIDDED RESPONSE Savannah has two sisters. One sister is 8 years older than her, and the other sister is 2 years younger than her. The product of Savannah’s sisters’ ages is 56. How old is Savannah?

SOLUTION: Let s = Savannah’s age. Then, s + 8 = the age of her older sister and s – 2 = the age of her younger sister.

The roots are –12 and 6. Savannah’s age cannot be negative. So, Savannah is 6 years old.

46. What is the product of a3b

5 and a

5b

2?

A a8b

7

B a2b

3

C a8b

3

D a2b

7

SOLUTION:

Choice A is the correct answer.

47. What is the solution set of x2 + 2x − 24 = 0?

F {−4, 6} G {−3, 8} H {3, −8} J {4, −6}

SOLUTION:

The solutions are –6 and 4. Choice J is the correct answer.

48. Which is the solution set of x ≥ −2? A

B

C

D

SOLUTION: Because x is greater than or equal to –2, there should be a closed circle at –2, and the line should be shaded to the right of –2. Choice C is the correct answer.

Factor each polynomial.

49. x2 − 9x + 14

SOLUTION:



Factors of 14 Sum –1, –14 –15 –2, –7 –9

50. n2 − 8n + 15

SOLUTION:



Factors of 15 Sum –1, –15 –16 –3, –5 –8

51. x2 − 5x − 24

SOLUTION:

In this trinomial, b = –5 and c = –24, so m + p is negative and mp is negative. Therefore, m and p must be opposite signs. List the factors of –24, and look for the pair of factors and identify the factors with a sum of –5.


Factors of –24 Sum 1, –24 –23 –1, 24 23 2, –12 –10 –2, 12 10 3, –8 –5 –3, 8 5 4, –6 –2 –4, 6 2

52. z2 + 15z + 36

SOLUTION:

In this trinomial, b = 15 and c = 36, so m + p is negative and mp is positive. Therefore, m and p must both be positive.List the factors of 36, and look for the pair of factors and identify the factors with a sum of 15.


Factors of 36 Sum 1, 36 37 2, 18 20 3, 12 15 4, 9 13 6, 6 12

53. r2 + 3r − 40

SOLUTION:

In this trinomial, b = 3 and c = –40, so m + p is positive and mp is negative. Therefore, m and p must be opposite signs. List the factors of –40, and look for the pair of factors and identify the factors with a sum of 3.


Factors of –40 Sum 1, –40 –39 –1, 40 39 2, –20 –18 –2, 20 18 4, –10 –6 –4, 10 6 5, –8 –3 –5, 8 3

54. v2 + 16v + 63

SOLUTION:

In this trinomial, b = 16 and c = 63, so m + p is positive and mp is positive. Therefore, m and p must be opposite signs. List the factors of 63, and look for the pair of factors and identify the factors with a sum of 16.


Factors of 63 Sum 1, 63 64 3, 21 24 7, 9 16

Solve each equation. Check your solutions.

55. a(a − 9) = 0

SOLUTION:


and


56. (2y + 6)(y − 1) = 0

SOLUTION:

The roots are –3 and 1. Check by substituting –3 and 1 in for y in the original equation.

and

The solutions are –3 and 1.

57. 10x2 − 20x = 0

SOLUTION:

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and


58. 8b2 − 12b = 0

SOLUTION:

The roots are 0 and . Check by substituting 0 and in for b in the original equation.

and


59. 15a2 = 60a

SOLUTION:


and


60. 33x2 = −22x

SOLUTION:

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and


61. ART A painter has 32 units of yellow dye and 54 units of blue dye to make two shades of green. The units needed to make a gallon of light green and a gallon of dark green are shown. Make a graph showing the numbers of gallons of the two greens she can make, and list three possible solutions.

SOLUTION: Let x = the number of gallons of light green and let y = the number of gallons of dark green. Then, 4x + y ≤ 32 and x + 6y ≤ 54. Graph 4x + y = 32 and x + 6y = 54. Both lines are included in the solution of the system and should be solid.

Some possible solutions are 2 light, 8 dark; 6 light, 8 dark; 7 light, 4 dark. They fall in the shaded region which represents the solution set.

Solve each compound inequality. Then graph the solution set.

62. k + 2 > 12 and k + 2 ≤ 18

SOLUTION:

The solution set is {k |10 < k ≤ 16}. To graph the solution set, graph k ≤ 16 and graph k > 10. Then find the intersection.

and

63. d − 4 > 3 or d − 4 ≤ 1

SOLUTION:

The solution set is {d|d ≤ 5 or d > 7}. Notice that the graphs do not intersect. To graph the solution set, graph d ≤ 5 and graph d > 7. Then find the union.

or

64. 3 < 2x − 3 < 15

SOLUTION:

The solution set is {x|3 < x < 9}.

To graph the solution set, graph x < 9 and graph x > 3. Then find the intersection.

and

65. 3t − 7 ≥ 5 and 2t + 6 ≤ 12

SOLUTION:

The solution set is empty. A number cannot be both greater than or equal to 4 and less than or equal to 3.

and

66. h − 10 < −21 or h + 3 < 2

SOLUTION:

The solution set is {h|h < −1}. To graph the solution set, graph h < –1. All numbers less than –1 will also be less than –11.

or

67. 4 < 2y − 2 < 10

SOLUTION:

The solution set is {y |3 < y < 6}.

To graph the solution set, graph y < 6 and graph y > 3. Then find the intersection.

and

68. FINANCIAL LITERACY A home security company provides security systems for $5 per week, plus an installation fee. The total cost for installation and 12 weeks of service is $210. Write the point-slope form of an equation to find the total fee y for any number of weeks x. What is the installation fee?

SOLUTION: The slope of the equation of the line that represents the weekly cost is 5. The total cost for 12 weeks of service including installation is $210. So, the line passes through the point (12, 210).

The point-slope form of an equation to find the total fee y for any number of weeks x is y − 210 = 5(x − 12). Solve the equation for y to find the flat fee for installation.

So, the flat fee for installation is $150.

Find the principal square root of each number.69. 16

SOLUTION:

70. 36

SOLUTION:

71. 64

SOLUTION:

72. 81

SOLUTION:

73. 121

SOLUTION:

74. 100

SOLUTION:

eSolutions Manual - Powered by Cognero Page 2

8-7 Solving ax^2 + bx + c = 0


1. 3x2 + 17x + 10

SOLUTION:



So, 3x2 + 17x + 10 = (3x + 2) (x +5).

Factors of 30 Sum 1, 30 31 2, 15 17 5, 6 30

2. 2x2 + 22x + 56

SOLUTION:


2x2




So, 2x2 + 22x + 56= 2(x + 4)(x + 7).

Factors of 28 Sum 1, 28 29 2, 14 16 4, 7 11

3. 5x2 − 3x + 4

SOLUTION:



Factors of 20 Sum 1, –20 –19 –1, 20 19 2, –10 –8 –2, 10 8 4, –5 –1 –4, 5 1

4. 3x2 − 11x − 20

SOLUTION:



So, 3x2 − 11x − 20 = (3x +4)(x – 5).

Factors of –60 Sum 1, –60 –59 –1, 60 59 2, –30 –28 –2, 30 28 3, –20 –17 –3, 20 17 4, –15 –11 –4, 15 11 5, –12 –7 –5, 12 7 6, –10 –4 –6, 10 4


5. 2x2 + 9x + 9 = 0







6. 3x2 + 17x + 20 = 0







7. 3x2 − 10x + 8 = 0







8. 2x2 − 17x + 30 = 0















10. 5x2 + 34x + 24

SOLUTION:



So, 5x2 + 34x + 24= (5x + 4)(x + 6).

Factors of 120 Sum 1, 120 121 2, 60 62 3, 40 43 4, 30 34 5, 24 29 6, 20 26 8, 15 23 10, 12 22

11. 2x2 + 19x + 24

SOLUTION:



So, 2x2 + 19x + 24= (2x + 3)(x + 8).

Factors of 48 Sum 1, 48 49 2, 24 26 3, 16 19 4, 12 16 6, 8 14

12. 4x2 + 22x + 10

SOLUTION:


4x2 + 22x + 10 = 2(2x




So, 4x2 + 22x + 10= 2(2x + 1)(x + 5).

Factors of 10 Sum 1, 10 11 2, 5 7

13. 4x2 + 38x + 70

SOLUTION:


4x2




So, 4x2 + 38x + 70= 2(2x + 5)(x + 7).

Factors of 70 Sum 1, 70 71 2, 35 37 5, 14 19 7, 10 17

14. 2x2 − 3x − 9

SOLUTION:



So, 2x2 − 3x − 9= (2x + 3)(x − 3).

Factors of –18 Sum 1, –18 –17 –1, 18 17 2, –9 –7 –2, 9 7 3, –6 –3 –3, 6 3

15. 4x2 − 13x + 10

SOLUTION:



So, 4x2 − 13x + 10= (4x − 5)(x − 2).

Factors of 40 Sum –1, –40 –41 –2, –20 –22 –4, –10 –14 –5, –8 –13

16. 2x2 + 3x + 6

SOLUTION:




Factors of 12 Sum 1, 12 13 2, 6 8 3, 4 7

17. 5x2 + 3x + 4

SOLUTION:




Factors of 20 Sum 1, 20 21 2, 10 12 4, 5 9

18. 12x2 + 69x + 45

SOLUTION:


12x2




So, 12x2 + 69x + 45= 3(4x + 3)(x + 5).

Factors of 60 Sum 1, 60 61 2, 30 32 3, 20 23 4, 15 19 5, 12 17 6, 10 60

19. 4x2 − 5x + 7

SOLUTION:




Factors of 28 Sum –1,–28 –29 –2, –14 –16 –4, –7 –11

20. 5x2 + 23x + 24

SOLUTION:



So, 5x2 + 23x + 24= (5x + 8)(x + 3).

Factors of 120 Sum 1, 120 121 2, 60 62 3, 40 43 4, 30 34 5, 24 29 6, 20 26 8, 15 23 10, 12 22

21. 3x2 − 8x + 15

SOLUTION:




Factors of 45 Sum –1,–45 –46 –3, –15 –18 –5, –9 –14


SOLUTION:







23. 2x2 + 9x − 18 = 0






24. 4x2 + 17x + 15 = 0






25. −3x2 + 26x = 16






26. −2x2 + 13x = 15





27. −3x2 + 5x = −2






28. −4x2 + 19x = −30







SOLUTION:




is 10 feet.


b.







SOLUTION:




32. −6x2 − 23x − 20





So, −6x2 − 23x − 20= −(2x + 5)(3x + 4).

Factors of 120 Sum 1, 120 121 2, 60 62 3, 40 43 4, 30 34 5, 24 29 6, 20 26 8, 15 23 10, 12 22

33. −4x2 − 15x − 14


−4x2 − 15x − 14 = –1(4x2 + 15x + 14)




So, −4x2 − 15x − 14 = −(x + 2)(4x + 7).

Factors of 56 Sum 1, 56 57 2, 28 30 4, 14 18 7, 8 115

34. −5x2 + 18x + 8


−5x2 + 18x + 8 = –1(5x2 – 18x – 8)




So, −5x2 + 18x + 8 = −(x – 4)(5x + 2).

Factors of –40 Sum –1, 40 39 1, –40 –39 –2, 20 18 2, –20 –18 –4, 10 6 4, –10 –6 –5, 8 3 5, –8 –3

35. −6x2 + 31x − 35


−6x2 + 31x − 35 = –1(6x2 – 31x + 35)




So, −6x2 + 31x − 35 = −(2x − 7)(3x − 5).

Factors of 210 Sum –1, –210 –211 –2, –105 –107 –3, –70 –73 –5, –42 –47 –6, –35 –41 –7, –30 –37 –10, –21 –31 –14, –15 –29

36. −4x2 + 5x − 12





Factors of 48 Sum –1, –48 –49 –2, –24 –26 –3, –16 –19 –4, –12 –16 –6, –8 –14

37. −12x2 + x + 20


−12x2 + x + 20 = –1(12x2 – x – 20)




So, −12x2 + x + 20 = −(4x + 5)(3x – 4).

Factors of –240 Sum –1, 240 239 1, –240 –239 –2, 120 118 2, –120 –118 –3, 80 77 3, –80 –77 –4, 60 56 4, –60 –56 –5, 48 43 5, –48 –43 –6, 40 34 6, –40 –34 –8, 30 22 8, –30 –22 –10, 24 14 10, –24 –14 –12, 20 8 12, –20 –8 –15, 16 1 15, –16 –1




SOLUTION:






SOLUTION: a.




b.



2.

c.




reasoning.




SOLUTION:


2.




SOLUTION: The values fo

esolutions manual - powered by cognero page 1 · &&6602'(/,1* ken throws the discus at a school...

Documents