the graphs coincide. therefore, the trinomial has been ... · therefore, the length and the width...

Factor each polynomial. Confirm your answers using a graphing calculator.

1. x2 + 14x + 24

SOLUTION:

In this trinomial, b = 14 and c = 24, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the factors of 24 and identify the factors with a sum of 14.

The correct factors are 2 and 12.

Confirm by graphing Y1 = x2 +14x + 24 and Y2 = (x + 2)(x + 12) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (x + 2)(x + 12).

Factors of 24 Sum 1, 24 25 2, 12 14 3, 8 11 4, 6 10

2. y2 − 7y − 30

SOLUTION:

In this trinomial, b = −7 and c = −30, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of −30 and identify the factors with a sum of −7.

The correct factors are 3 and −10.

Confirm by graphing Y1 = y2 – 7y – 30 + and Y2 = (x – 10)(x + 3) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (x – 10)(x + 3).

Factors of −30 Sum 1, −30 −29 2, −15 −13 3, −10 −7 5, −6 −1 6, −5 1 10, −3 7 15, −2 13 30, −1 29

3. n2 + 4n − 21

SOLUTION:

In this trinomial, b = 4 and c = −21, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of −21 and identify the factors with a sum of 4.


Confirm by graphing Y1 = n2 + 4n– 21 + and Y2 = (n – 3)(n + 7) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (n – 3)(n + 7).

Factors of −21 Sum 1, −21 −20 3, −7 −4 7, −3 4 21, −1 20

4. m2 − 15m + 50

SOLUTION:

In this trinomial, b = –15 and c = 50, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the factors of 50 and identify the factors with a sum of –15.

The correct factors are –5 and –10.

Confirm by graphing Y1 = m2 – 15m + 50 + and Y2 = (m – 5)(m – 10) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (m – 5)(m – 10).

Factors of 50 Sum –1, –50 –51 –2, –25 –27 –5, –10 –15

Solve each equation. Check your solutions.

5. x2 − 4x − 21 = 0

SOLUTION: List the factors of 21 and identify the factors with a sum of –4.

The roots are –3 and 7. Check by substituting –3 and 7 in for x in the original equation.

and

The solutions are –3 and 7.

Factors of 21 Sum 1, –21 –20 –1, 21 19 3, –7 –4 –3, 7 4

6. n2 − 3n + 2 = 0


The roots are 1 and 2. Check by substituting 1 and 2 in for n in the original equation.

and

The solutions are 1 and 2.

Factors of 2 Sum 1, 2 3 –1, –2 –3

7. x2 − 15x + 54 = 0


The roots are 9 and 6. Check by substituting 9 and 6 in for x in the original equation.

and


Factors of 54 Sum 1, 54 55 –1, –54 –55

2, 27 29 –2, –27 –29

3, 18 21 –3, –18 –21

6, 9 15 –6, –9 –15

8. x2 + 12x = −32

SOLUTION: Rewrite the equation with 0 on the right side.

List the factors of 32 and identify the factors with a sum of 12.

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and -4.

Factors of 32 Sum 1, 32 33 2, 16 18 4, 8 12

9. x2 − x − 72 = 0



and


Factors of 72 Sum –1, 72 71 1, –72 –71 –2, 36 34 2, –36 –34 –3, 24 21 3, –24 –21 –4, 18 14 4, –18 –14 –6, 12 66,–12 –6–8, 9 1 8, –9 –1

10. x2 − 10x = −24


. List the factors of 24 and identify the factors with a sum of –10.


and


Factors of 24 Sum –1, –24 –25 –2, –12 –14 –3, –8 –11 –4, –6 –10

11. FRAMING Tina bought a frame for a photo, but the photo is too big for the frame. Tina needs to reduce the width and length of the photo by the same amount. The area of the photo should be reduced to half the original area. If the original photo is 12 inches by 16 inches, what will be the dimensions of the smaller photo?

SOLUTION: Let x be the amount that Tina should reduce the photo. So the dimensions are now (12 – x)(16 – x). The original area was 12(16) = 192 square inches. Since the area is to be reduced by half, the new area will be 96 square inches.Now solve the equation.

The answer x = 24 does not make sense because it would result in a new length of 12 – 24, or –12 inches. Therefore, the length and the width of the photo must both be reduced by 4 inches. So, the new dimensions are 8 inches by 12 inches.


12. x2 + 17x + 42

SOLUTION:



Confirm by graphing Y1 = x2 + 17x + 42 and Y2 = (x + 3)(x + 14) on the same screen.


Factors of 42 Sum 1, 42 43 2, 21 23 3, 14 17 6, 7 13

13. y2 − 17y + 72

SOLUTION:

In this trinomial, b = −17 and c = 72, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the factors of 72 and identify the factors with a sum of −17.

The correct factors are −8 and −9.

Confirm by graphing Y1 = y2 − 17y + 72 and Y2 = (y – 8)(y – 9) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (y – 8)(y – 9).

Factors of 72 Sum −1, −72 −73 −2, −36 −38 −3, −24 −27 −4, −18 −22 −6, −12 −18 −8, −9 −17

14. a2 + 8a − 48

SOLUTION:

In this trinomial, b = 8 and c = −48, so m + p is positive and mp is negative. Therefore, m and p must have opposite signs. List the factors of −48 and identify the factors with a sum of 8.

The correct factors are −4 and 12.

Confirm by graphing Y1 = a2 + 8a − 48 and Y2 = (a – 4)(a + 12) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (a – 4)(a + 12).

Factors of −48 Sum −1, 48 47 −2, 24 22 −3, 16 13 −4, 12 8 −6, 8 2 −8, 6 −2 −12, 4 −8 −16, 3 −13 −24, 2 −22 −48, 1 −47

15. n2 − 2n − 35

SOLUTION:

In this trinomial, b = −2 and c = −35, so m + p is negative and mp is negative. Therefore, m and p must have oppositesigns. List the factors of −35 and identify the factors with a sum of −2.


Confirm by graphing Y1 = n2 – 2n – 35 and Y2 = (n – 7)(n + 5) on the same screen.


Factors of −35 Sum −1, 35 34 −5, 7 2 −7, 5 −2 −35, 1 −34

16. 44 + 15h + h2

SOLUTION:

First rearrange the polynomial in decreasing order, h2 + 15h + 44.



Confirm by graphing Y1 = 44 + 15h + h2 and Y2 = (h + 4)(h + 11) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (h + 4)(h + 11).

Factors of 44 Sum 1, 44 45 2, 22 24 4, 11 15

17. 40 − 22x + x2

SOLUTION:

First rearrange the polynomial in decreasing order, x2 – 22x + 40.



Confirm by graphing Y1 = 40 – 22x + x2 and Y2 = (x – 2)(x – 20) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (x – 2)(x – 20).

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

18. −24 − 10x + x2

SOLUTION:

First rearrange the polynomial in decreasing order x2 – 10x – 24.

In this trinomial, b = –10 and c = –24, so m + p is negative and mp is negative. Therefore, m and p must have opposite signs. List the factors of –24 and identify the factors with a sum of –10.

The correct factors are –12 and 2.

Confirm by graphing Y1 = –24 – 10x + x2 and Y2 = (x + 2)(x – 12) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (x + 2)(x – 12).

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

19. −42 − m + m2

SOLUTION:

First rearrange the polynomial in decreasing order m2 – m – 42.

In this trinomial, b = –1 and c = –42, so m + p is negative and mp is negative. Therefore, m and p must have oppositesigns. List the factors of –42 and identify the factors with a sum of –1.


Confirm by graphing Y1 = –42 – m + m2 and Y2 = (m + 6)(m – 7) on the same screen.

The graphs coincide. Therefore, the trinomial has been factored correctly. √ The factors are (m + 6)(m – 7).

Factors of –42 Sum –1, 42 41 –2, 21 19 –3, 14 11 –6, 7 1 –7, 6 –1 –14, 3 –11 –21, 2 –19 –42, 1 –41


20. x2 − 7x + 12 = 0



and


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

21. y2 + y = 20


List the factors of –20 and identify the factors with a sum of 1.

The roots are –5 and 4. Check by substituting –5 and 4 in for y in the original equation.

and


Factors of –20 Sum 1, –20 –19 –1, 20 19 2,–10 –8 –2, 10 84, –5 –1

–4, 5 1

22. x2 − 6x = 27


List the factors of –27 and identify the factors with a sum of –6.


and


Factors of –27 Sum 1, –27 –26 –1, 27 26 3, –9 –6 –3, 9 6

23. a2 + 11a = −18



The roots are –2 and –9. Check by substituting –2 and –9 in for a in the original equation.

and

The solutions are –2 and –9.

Factors of 18 Sum 1, 18 19 2, 9 11 3, 6 19

24. c2 + 10c + 9 = 0

SOLUTION: List the factors of 9 and identify the factors with a sum of 10.

The roots are –1 and –9. Check by substituting –1 and –9 in for c in the original equation.

and


Factors of 9 Sum 1, 9 10 3, 3 6

25. x2 − 18x = −32


List the factors of 32 and identify the factors with a sum of –18.


and


Factors of 32 Sum –1, –32 –33 –2, –16 –18 –4, –8 –12

26. n2 − 120 = 7n


List the factors of −120 and identify the factors with a sum of −7.

The roots are 15 and −8. Check by substituting 15 and −8 in for n in the original equation.

and

The solutions are −8 and 15.

Factors of −120 Sum −1, 120 119 1, −120 −119 −2, 60 58 2, −60 −58 −3, 40 373, −40 −37

−4, 30 264, −30 −26

−5, 24 195, −24 −19

−6, 20 146, −20 −14

−8, 15 78, −15 −7

−12, 10 −212, −10 2

27. d2 + 56 = −18d



The roots are −4 and −14. Check by substituting −4 and −14 in for d in the original equation.

and

The solutions are −4 and −14.

Factors of 56 Sum 1, 56 57 2, 38 40 4, 14 18 7, 8 15

28. y2 − 90 = 13y



The roots are −5 and 18. Check by substituting −5 and 18 in for y in the original equation.

and


Factors of 90 Sum 1, −90 −89 −1, 90 89 2, −45 −43 −2, 45 43 3, −30 −27

−3, 30 275 ,−18 −13

−5, 18 136, −15 −9

−6, 15 99, −10 −1

−9, 10 1

29. h2 + 48 = 16h



The roots are 4 and 12. Check by substituting 4 and 12 in for h in the original equation.

and


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

30. GEOMETRY A triangle has an area of 36 square feet. If the height of the triangle is 6 feet more than its base, what are its height and base?

SOLUTION: Let b represent the base. Then b + 6 is the height.


However, the base cannot be negative, so the base is 6 feet. The height is b + 6 = 6 + 6 = 12 feet.

Factors of –72 Sum 1, –72 –71 –1, 72 71 2, –36 –34 –2, 36 34 3, –24 –21

–3, 24 214, –18 –14

–4, 18 146, –12 –6

–6, 12 68, –9 –1

–8, 9 1

31. GEOMETRY A rectangle has an area represented by x2 − 4x − 12 square feet. If the length is x + 2 feet, what is

the width of the rectangle?

SOLUTION:

The area of the rectangle is x2 − 4x − 12.


Then area of the rectangle is (x + 2)(x – 6). Area is found by multiplying the length by the width. Because the lengthis x + 2, the width must be x − 6.

Factors of −12 Sum 1, −12 −11 −1, 12 11 2, −6 −4 −2, 6 4 3, −4 −1

−3, 4 1

32. SOCCER The width of a high school soccer field is 45 yards shorter than its length.

a. Define a variable, and write an expression for the area of the field. b. The area of the field is 9000 square yards. Find the dimensions.

SOLUTION: a. Let = length. The area of the field is the length times the width, or ( − 45). b.

Then length cannot be negative, so it is 120 yd and the width is 120 – 45, or 75 yards.

CCSS STRUCTURE Factor each polynomial.

33. q2 + 11qr + 18r

2

SOLUTION:


The correct factors are 2 and 9. The trinomial has two variables q and r. The first term in each binomial will have q's, the second term will have the r along with the factors.

Factors of 18 Sum 1, 18 19 2, 9 11 3, 6 9

34. x2 − 14xy − 51y2

SOLUTION:


The correct factors are 3 and −17. The trinomial has two variables x and y . The first term in each binomial will have x 's, the second term will have they along with the factors.

Factors of −51 Sum −1, 51 50 −3, 17 14 −17, 3 −14 −51, 1 −50

35. x2 − 6xy + 5y

2

SOLUTION:

In this trinomial, b = −6 and c = 5, so m + p is negative and mp is positive. Therefore, m and p must both be negative.List the negative factors of 5 and identify the factors with a sum of −6.

The correct factors are −1 and −6. The trinomial has two variables x and y . The first term in each binomial will have x 's, the second term will have they along with the factors.

Factors of 5 Sum −1, −5 −6

36. a2 + 10ab − 39b2

SOLUTION:


The correct factors are −3 and 13. The trinomial has two variables a and b. The first term in each binomial will have as, the second term will have the b along with the factors.

Factors of −39 Sum 1, −39 −38 3, −13 −10 −3, 13 10 −39, 1 −38

37. SWIMMING The length of a rectangular swimming pool is 20 feet greater than its width. The area of the pool is 525 square feet. a. Define a variable and write an equation for the area of the pool. b. Solve the equation. c. Interpret the solutions. Do both solutions make sense? Explain.

SOLUTION: a. Sample answer: Let w = width. Then the length is 20 feet greater than its width, so = w + 20. The area, which is 525 square feet, is found by multiplying the length times the width. Therefore the area is (w + 20)w = 525. b.

List the factors of −525 and identify the factors with a sum of 20.

The width cannot be negative, therefore it is 15 feet. The length is 15 + 20, 35 feet. c. The solution of 15 means that the width is 15 ft. The solution −35 does not make sense because the width cannot be negative.

Factors of −525 Sum 1, −525 −524 −1, 525 524 5, −105 −100 −5, 105 100 7, −75 −68

−7, 75 6815, −35 −20

−15, 35 2021, −25 −4

−21, 25 4

GEOMETRY Find an expression for the perimeter of a rectangle with the given area.

38. A = x2 + 24x − 81

SOLUTION:

The area of the rectangle is x2 + 24x − 81.

In this trinomial, b = 24 and c = 24, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the factors of − 81 and identify the factors with a sum of 24.

Then area of the rectangle of x2 + 24x − 81 can be factored to (x + 27)(x – 3).

Area is found by multiplying the length by the width, so the length is x + 27 and the width is x − 3.

So, an expression for the perimeter of the rectangle is 4x + 48.

Factors of − 81 Sum 1,−81 −80 −1, 81 80 3, −27 −24 −3, 27 24 9,−9 0

39. A = x2 + 13x − 90

SOLUTION:


In this trinomial, b = 13 and c = − 90, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the factors of − 90 and identify the factors with a sum of 13.

The area of the rectangle x2 + 13x − 90 factors to (x + 18)(x – 5).



Factors of − 90 Sum 1, −90 −89 −1, 90 89 2, −45 −43 −2, 45 43 3, −30 −27

−3, 30 275, −18 −13

−5, 18 136, −15 −9

−6, 15 99, −10 −1

−9, 10 1

40. MULTIPLE REPRESENTATIONS In this problem, you will explore factoring when the leading coefficient is not 1. a. TABULAR Copy and complete the table below.

b. ANALYTICAL How are m and p related to a and c? c. ANALYTICAL How are m and p related to b?

d. VERBAL Describe a process you can use for factoring a polynomial of the form ax2 + bx + c.

SOLUTION: a. Use FOIL or the Distributive Property to find the product of the binomials.

b. ac is the product of the coefficients of the first and last terms. mp is the product of the coefficients of the middle two terms when foiling. Looking at the last two columns and you will see that mp = ac. c. Look at the second and third columns and you will see that m + p = b. The coefficients of the middle terms equalb. d. When factoring trinomials , we look for two integers, m and p , for which mp = ac and m + p = b.

41. ERROR ANALYSIS Jerome and Charles have factored x2 + 6x − 16. Is either of them correct? Explain your reasoning.

SOLUTION: Charles is correct. In this trinomial, b = 6 and c = − 16, so m + p is positive and mp is positive. Therefore, m and p must both be positive.List the factors of − 16 and identify the factors with a sum of − 6.

The correct factors are −2, 8. Jerome’s answer once multiplied is x2 − 6x − 16. The middle term should be positive.

Factors of − 16 Sum 1, −16 −15 −1, 16 15 2, −8 −6 −2, 8 6 4, −4 0

CCSS ARGUMENTS Find all values of k so that each polynomial can be factored using integers.

42. x2 + kx − 19

SOLUTION:

In this trinomial, b = k , c = −19, and mp is negative. Therefore, m + p must equal the sum of the factors of −19. List the factors of −19. The sum of m + p will equal k .

Therefore, k could have values of 18 or −18.

Factors of −19 Sum of m + p −1, 19 18 −19, 1 −18

43. x2 + kx + 14

SOLUTION:

In this trinomial, b = k , c = 14 and mp is positive. Therefore, m + p must equal the sum of the factors of 14. List the factors of 14. The sum of m + p will equal k .

Therefore, k could have values of 9, 15, −9 or −15.

Factors of 14 Sum of m + p 1, 14 15 2, 7 9

−1, −14 −15 −2, −7 −9

44. x2 − 8x + k , k > 0

SOLUTION:

In this trinomial, b = −8, c = k and m + p is negative. Therefore, mp must equal the product of the two numbers that add to −8. Because k > 0, this means that both m and p are negative. List the numbers that add to −8. The product ofmp will equal k .

Therefore, k could have values of 7, 12, 15, or 16.

Two numbers that add

to −8 Product of mp

−1, −7 7 −2, −6 12 −3, −5 15 −4, −4 16

45. x2 − 5x + k , k > 0

SOLUTION:


Therefore, k could have values of 4 or 6.



−1, −4 4 −2, −3 6

46. REASONING For any factorable trinomial, x2 + bx + c, will the absolute value of b sometimes, always, or never be less than the absolute value of c? Explain.

SOLUTION:

The absolute value of b will sometimes be less than the absolute value of c. Sample answer: The trinomial x2 + 10x +

9 = (x + 1)(x + 9) and 10 > 9. The trinomial x2 + 7x + 10 = (x + 2)(x + 5) and 7 < 10.

47. OPEN ENDED Give an example of a trinomial that can be factored using the factoring techniques presented in thislesson. Then factor the trinomial.

SOLUTION:

Students’ answers will vary. Sample answer: x2 + 19x − 20

In this trinomial, b = 19 and c = −20, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of −20, and look for the pair of factors with a sum of 19.


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

48. CHALLENGE Factor (4y − 5)2 + 3(4y − 5) − 70.

SOLUTION:

The trinomial is written in x2 + bx + c = 0 form.We can substitute x for 4y – 5 to get x

2 + 3x – 70 = 0.



Factors of −70 Sum –1, 70 69 −2, 35 33 −5, 14 9 −7, 10 3 −10, 7 −3 −14, 5 −9 −35, 2 −33 −70, 1 −69

49. WRITING IN MATH Explain how to factor trinomials of the form x2 + bx + c and how to determine the signs of the factors of c.

SOLUTION: Find factors m and p such that m + p = b and mp = c. If b and c are positive, then m and p are positive.

For example: If b is negative and c is positive, then m and p are negative.

For example: When c is negative, m and p have different signs and the factor with the greatest absolute value has the same sign asb.

For example: and

50. Which inequality is shown in the graph?

A

B

C

D

SOLUTION: The line is dashed which means that choices A and D can be eliminated. So check (0, 0) in the inequalities given in choices B and C to determine which is the correct choice.

So, the correct choice is C.

True

False

51. SHORT RESPONSE Olivia must earn more than $254 from selling candy bars in order to go on a trip with the National Honor Society. If each candy bar is sold for $1.25, what is the fewest candy bars she must sell?

SOLUTION: Let x represent the number of candy bars Olivia must sell.

So, Olivia must sell at least 204 candy bars to go on the trip.

52. GEOMETRY Which expression represents the length of the rectangle?

F x + 5 G x + 6 H x − 6 J x − 5

SOLUTION:


In this trinomial, b = − 3and c = − 18, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the factors of − 18 and identify the factors with a sum of − 3.

The area of the rectangle x2 − 3x − 18 factors to (x + 3)(x – 6). Because the width is x + 3, the length must be x − 6. So, the correct choice is H.

Factors of − 18 Sum 1,−18 −17 −1, 18 17 2, −9 −6 −2, 9 6 3, −6 −3

−3, 6 3

53. The difference of 21 and a number n is 6. Which equation shows the relationship? A 21 − n = 6 B 21 + n = 6 C 21n = 6 D 6n = −21

SOLUTION: The phrase “is 6” means “= 6”, so choice D can be eliminated. The phrase “difference of” means “subtraction”, so A is the correct choice.

Factor each polynomial.

54. 10a2 + 40a

SOLUTION:

The greatest common factor is or 10a.

10a2 + 40a = 10a(a + 4)

55. 11x + 44x2y

SOLUTION:

The greatest common factor is or 11x.

11x + 44x2y = 11x(1 + 4xy)

56. 2m3p

2 − 16mp2 + 8mp

SOLUTION:

The greatest common factor is or 2mp.

2m3p

2 − 16mp2 + 8mp = 2mp(m2p − 8p + 4)

57. 2ax + 6xc + ba + 3bc

SOLUTION: Factor by grouping.

58. 8ac − 2ad + 4bc − bd


59. x2 − xy − xy + y

2


60. Write a polynomial that represents the area of the shaded region in the figure.

SOLUTION:

Use elimination to solve each system of equations.

61.

SOLUTION: Because –x and x have opposite coefficients, add the equations.

Now, substitute 13 for y in either equation to find the value of x.

Check the solution in each equation.

Therefore, the solution is (4, 13).

62.

SOLUTION: Because 5a and –5a have opposite coefficients, add the equations.

Now, substitute –3 for b in either equation to find the value of a.


Therefore, the solution is (2, –3).

63.

SOLUTION:

Because d and −d have opposite coefficients, add the equations.

Now, substitute 3 for c in either equation to find the value of d.



64.

SOLUTION:

Because 2y and −2y have opposite coefficients, add the equations.

Now, substitute 2 for x in either equation to find the value of y .



65. LANDSCAPING Kendrick is planning a circular flower garden with a low fence around the border. He has 38 feet of fence. What is the radius of the largest garden he can make? (Hint: C = 2πr)

SOLUTION: If he has 38 feet of fence, use the formula, C = 2πr, with C = 38.

So, the largest radius he can make is about 6 feet.


66. 6mx − 4m + 3rx − 2r

SOLUTION:

67. 3ax − 6bx + 8b − 4a

SOLUTION:

68. 2d2g + 2fg + 4d2h + 4fh

SOLUTION:

eSolutions Manual - Powered by Cognero Page 1

8-6 Solving x^2 + bx + c = 0


1. x2 + 14x + 24

SOLUTION:





Factors of 24 Sum 1, 24 25 2, 12 14 3, 8 11 4, 6 10

2. y2 − 7y − 30

SOLUTION:





Factors of −30 Sum 1, −30 −29 2, −15 −13 3, −10 −7 5, −6 −1 6, −5 1 10, −3 7 15, −2 13 30, −1 29

3. n2 + 4n − 21

SOLUTION:





Factors of −21 Sum 1, −21 −20 3, −7 −4 7, −3 4 21, −1 20

4. m2 − 15m + 50

SOLUTION:





Factors of 50 Sum –1, –50 –51 –2, –25 –27 –5, –10 –15


5. x2 − 4x − 21 = 0



and


Factors of 21 Sum 1, –21 –20 –1, 21 19 3, –7 –4 –3, 7 4

6. n2 − 3n + 2 = 0



and


Factors of 2 Sum 1, 2 3 –1, –2 –3

7. x2 − 15x + 54 = 0



and


Factors of 54 Sum 1, 54 55 –1, –54 –55

2, 27 29 –2, –27 –29

3, 18 21 –3, –18 –21

6, 9 15 –6, –9 –15

8. x2 + 12x = −32




and


Factors of 32 Sum 1, 32 33 2, 16 18 4, 8 12

9. x2 − x − 72 = 0



and


Factors of 72 Sum –1, 72 71 1, –72 –71 –2, 36 34 2, –36 –34 –3, 24 21 3, –24 –21 –4, 18 14 4, –18 –14 –6, 12 66,–12 –6–8, 9 1 8, –9 –1

10. x2 − 10x = −24




and


Factors of 24 Sum –1, –24 –25 –2, –12 –14 –3, –8 –11 –4, –6 –10





12. x2 + 17x + 42

SOLUTION:





Factors of 42 Sum 1, 42 43 2, 21 23 3, 14 17 6, 7 13

13. y2 − 17y + 72

SOLUTION:





Factors of 72 Sum −1, −72 −73 −2, −36 −38 −3, −24 −27 −4, −18 −22 −6, −12 −18 −8, −9 −17

14. a2 + 8a − 48

SOLUTION:





Factors of −48 Sum −1, 48 47 −2, 24 22 −3, 16 13 −4, 12 8 −6, 8 2 −8, 6 −2 −12, 4 −8 −16, 3 −13 −24, 2 −22 −48, 1 −47

15. n2 − 2n − 35

SOLUTION:





Factors of −35 Sum −1, 35 34 −5, 7 2 −7, 5 −2 −35, 1 −34

16. 44 + 15h + h2

SOLUTION:






Factors of 44 Sum 1, 44 45 2, 22 24 4, 11 15

17. 40 − 22x + x2

SOLUTION:






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

18. −24 − 10x + x2

SOLUTION:






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

19. −42 − m + m2

SOLUTION:






Factors of –42 Sum –1, 42 41 –2, 21 19 –3, 14 11 –6, 7 1 –7, 6 –1 –14, 3 –11 –21, 2 –19 –42, 1 –41


20. x2 − 7x + 12 = 0



and


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

21. y2 + y = 20




and


Factors of –20 Sum 1, –20 –19 –1, 20 19 2,–10 –8 –2, 10 84, –5 –1

–4, 5 1

22. x2 − 6x = 27




and


Factors of –27 Sum 1, –27 –26 –1, 27 26 3, –9 –6 –3, 9 6

23. a2 + 11a = −18




and


Factors of 18 Sum 1, 18 19 2, 9 11 3, 6 19

24. c2 + 10c + 9 = 0



and


Factors of 9 Sum 1, 9 10 3, 3 6

25. x2 − 18x = −32




and


Factors of 32 Sum –1, –32 –33 –2, –16 –18 –4, –8 –12

26. n2 − 120 = 7n




and


Factors of −120 Sum −1, 120 119 1, −120 −119 −2, 60 58 2, −60 −58 −3, 40 373, −40 −37

−4, 30 264, −30 −26

−5, 24 195, −24 −19

−6, 20 146, −20 −14

−8, 15 78, −15 −7

−12, 10 −212, −10 2

27. d2 + 56 = −18d




and


Factors of 56 Sum 1, 56 57 2, 38 40 4, 14 18 7, 8 15

28. y2 − 90 = 13y




and


Factors of 90 Sum 1, −90 −89 −1, 90 89 2, −45 −43 −2, 45 43 3, −30 −27

−3, 30 275 ,−18 −13

−5, 18 136, −15 −9

−6, 15 99, −10 −1

−9, 10 1

29. h2 + 48 = 16h




and


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





Factors of –72 Sum 1, –72 –71 –1, 72 71 2, –36 –34 –2, 36 34 3, –24 –21

–3, 24 214, –18 –14

–4, 18 146, –12 –6

–6, 12 68, –9 –1

–8, 9 1



SOLUTION:




Factors of −12 Sum 1, −12 −11 −1, 12 11 2, −6 −4 −2, 6 4 3, −4 −1

−3, 4 1






33. q2 + 11qr + 18r

2

SOLUTION:



Factors of 18 Sum 1, 18 19 2, 9 11 3, 6 9

34. x2 − 14xy − 51y2

SOLUTION:



Factors of −51 Sum −1, 51 50 −3, 17 14 −17, 3 −14 −51, 1 −50

35. x2 − 6xy + 5y

2

SOLUTION:




36. a2 + 10ab − 39b2

SOLUTION:



Factors of −39 Sum 1, −39 −38 3, −13 −10 −3, 13 10 −39, 1 −38





Factors of −525 Sum 1, −525 −524 −1, 525 524 5, −105 −100 −5, 105 100 7, −75 −68

−7, 75 6815, −35 −20

−15, 35 2021, −25 −4

−21, 25 4


38. A = x2 + 24x − 81

SOLUTION:






Factors of − 81 Sum 1,−81 −80 −1, 81 80 3, −27 −24 −3, 27 24 9,−9 0

39. A = x2 + 13x − 90

SOLUTION:






Factors of − 90 Sum 1, −90 −89 −1, 90 89 2, −45 −43 −2, 45 43 3, −30 −27

−3, 30 275, −18 −13

−5, 18 136, −15 −9

−6, 15 99, −10 −1

−9, 10 1









Factors of − 16 Sum 1, −16 −15 −1, 16 15 2, −8 −6 −2, 8 6 4, −4 0


42. x2 + kx − 19

SOLUTION:




43. x2 + kx + 14

SOLUTION:




−1, −14 −15 −2, −7 −9

44. x2 − 8x + k , k > 0

SOLUTION:





−1, −7 7 −2, −6 12 −3, −5 15 −4, −4 16

45. x2 − 5x + k , k > 0

SOLUTION:





−1, −4 4 −2, −3 6


SOLUTION:




SOLUTION:




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


SOLUTION:

The trinomial is written in x2 + bx + c = 0 form.We can substitute x for 4y – 5 to get x

2 + 3x – 70 = 0.



Factors of −70 Sum –1, 70 69 −2, 35 33 −5, 14 9 −7, 10 3 −10, 7 −3 −14, 5 −9 −35, 2 −33 −70, 1 −69

49. WRITING IN MATH Explain how to factor trinomials of the form x2 + bx + c and how to determine the signs of the factors of c.

SOLUTION: Find factors m and p such that m + p = b and mp = c. If b and c are positive, then m and p are positive.

For example: If b is negative and c is positive, then m and p are negative.

For example: When c is negative, m and p have different signs and the factor with the greatest absolute value has the same sign asb.

For example: and

50. Which inequality is shown in the graph?

A

B

C

D

SOLUTION: The line is dashed which means that choices A and D can be eliminated. So check (0, 0) in the inequalities given in choices B and C to determine which is the correct choice.

So, the correct choice is C.

True

False

51. SHORT RESPONSE Olivia must earn more than $254 from selling candy bars in order to go on a trip with the National Honor Society. If each candy bar is sold for $1.25, what is the fewest candy bars she must sell?

SOLUTION: Let x represent the number of candy bars Olivia must sell.

So, Olivia must sell at least 204 candy bars to go on the trip.

52. GEOMETRY Which expression represents the length of the rectangle?

F x + 5 G x + 6 H x − 6 J x − 5

SOLUTION:


In this trinomial, b = − 3and c = − 18, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the factors of − 18 and identify the factors with a sum of − 3.

The area of the rectangle x2 − 3x − 18 factors to (x + 3)(x – 6). Because the width is x + 3, the length must be x − 6. So, the correct choice is H.

Factors of − 18 Sum 1,−18 −17 −1, 18 17 2, −9 −6 −2, 9 6 3, −6 −3

−3, 6 3

53. The difference of 21 and a number n is 6. Which equation shows the relationship? A 21 − n = 6 B 21 + n = 6 C 21n = 6 D 6n = −21

SOLUTION: The phrase “is 6” means “= 6”, so choice D can be eliminated. The phrase “difference of” means “subtraction”, so A is the correct choice.


54. 10a2 + 40a

SOLUTION:

The greatest common factor is or 10a.

10a2 + 40a = 10a(a + 4)

55. 11x + 44x2y

SOLUTION:

The greatest common factor is or 11x.

11x + 44x2y = 11x(1 + 4xy)

56. 2m3p

2 − 16mp2 + 8mp

SOLUTION:

The greatest common factor is or 2mp.

2m3p

2 − 16mp2 + 8mp = 2mp(m2p − 8p + 4)

57. 2ax + 6xc + ba + 3bc


58. 8ac − 2ad + 4bc − bd


59. x2 − xy − xy + y

2


60. Write a polynomial that represents the area of the shaded region in the figure.

SOLUTION:

Use elimination to solve each system of equations.

61.

SOLUTION: Because –x and x have opposite coefficients, add the equations.

Now, substitute 13 for y in either equation to find the value of x.



62.

SOLUTION: Because 5a and –5a have opposite coefficients, add the equations.

Now, substitute –3 for b in either equation to find the value of a.


Therefore, the solution is (2, –3).

63.

SOLUTION:

Because d and −d have opposite coefficients, add the equations.

Now, substitute 3 for c in either equation to find the value of d.



64.

SOLUTION:

Because 2y and −2y have opposite coefficients, add the equations.

Now, substitute 2 for x in either equation to find the value of y .



65. LANDSCAPING Kendrick is planning a circular flower garden with a low fence around the border. He has 38 feet of fence. What is the radius of the largest garden he can make? (Hint: C = 2πr)

SOLUTION: If he has 38 feet of fence, use the formula, C = 2πr, with C = 38.

So, the largest radius he can make is about 6 feet.


66. 6mx − 4m + 3rx − 2r

SOLUTION:

67. 3ax − 6bx + 8b − 4a

SOLUTION:

68. 2d2g + 2fg + 4d2h + 4fh

SOLUTION:

eSolutions Manual - Powered by Cognero Page 2

8-6 Solving x^2 + bx + c = 0


1. x2 + 14x + 24

SOLUTION:





Factors of 24 Sum 1, 24 25 2, 12 14 3, 8 11 4, 6 10

2. y2 − 7y − 30

SOLUTION:





Factors of −30 Sum 1, −30 −29 2, −15 −13 3, −10 −7 5, −6 −1 6, −5 1 10, −3 7 15, −2 13 30, −1 29

3. n2 + 4n − 21

SOLUTION:





Factors of −21 Sum 1, −21 −20 3, −7 −4 7, −3 4 21, −1 20

4. m2 − 15m + 50

SOLUTION:





Factors of 50 Sum –1, –50 –51 –2, –25 –27 –5, –10 –15


5. x2 − 4x − 21 = 0



and


Factors of 21 Sum 1, –21 –20 –1, 21 19 3, –7 –4 –3, 7 4

6. n2 − 3n + 2 = 0



and


Factors of 2 Sum 1, 2 3 –1, –2 –3

7. x2 − 15x + 54 = 0



and


Factors of 54 Sum 1, 54 55 –1, –54 –55

2, 27 29 –2, –27 –29

3, 18 21 –3, –18 –21

6, 9 15 –6, –9 –15

8. x2 + 12x = −32




and


Factors of 32 Sum 1, 32 33 2, 16 18 4, 8 12

9. x2 − x − 72 = 0



and


Factors of 72 Sum –1, 72 71 1, –72 –71 –2, 36 34 2, –36 –34 –3, 24 21 3, –24 –21 –4, 18 14 4, –18 –14 –6, 12 66,–12 –6–8, 9 1 8, –9 –1

10. x2 − 10x = −24




and


Factors of 24 Sum –1, –24 –25 –2, –12 –14 –3, –8 –11 –4, –6 –10





12. x2 + 17x + 42

SOLUTION:





Factors of 42 Sum 1, 42 43 2, 21 23 3, 14 17 6, 7 13

13. y2 − 17y + 72

SOLUTION:





Factors of 72 Sum −1, −72 −73 −2, −36 −38 −3, −24 −27 −4, −18 −22 −6, −12 −18 −8, −9 −17

14. a2 + 8a − 48

SOLUTION:





Factors of −48 Sum −1, 48 47 −2, 24 22 −3, 16 13 −4, 12 8 −6, 8 2 −8, 6 −2 −12, 4 −8 −16, 3 −13 −24, 2 −22 −48, 1 −47

15. n2 − 2n − 35

SOLUTION:





Factors of −35 Sum −1, 35 34 −5, 7 2 −7, 5 −2 −35, 1 −34

16. 44 + 15h + h2

SOLUTION:






Factors of 44 Sum 1, 44 45 2, 22 24 4, 11 15

17. 40 − 22x + x2

SOLUTION:






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

18. −24 − 10x + x2

SOLUTION:






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

19. −42 − m + m2

SOLUTION:






Factors of –42 Sum –1, 42 41 –2, 21 19 –3, 14 11 –6, 7 1 –7, 6 –1 –14, 3 –11 –21, 2 –19 –42, 1 –41


20. x2 − 7x + 12 = 0



and


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

21. y2 + y = 20




and


Factors of –20 Sum 1, –20 –19 –1, 20 19 2,–10 –8 –2, 10 84, –5 –1

–4, 5 1

22. x2 − 6x = 27




and


Factors of –27 Sum 1, –27 –26 –1, 27 26 3, –9 –6 –3, 9 6

23. a2 + 11a = −18




and


Factors of 18 Sum 1, 18 19 2, 9 11 3, 6 19

24. c2 + 10c + 9 = 0



and


Factors of 9 Sum 1, 9 10 3, 3 6

25. x2 − 18x = −32




and


Factors of 32 Sum –1, –32 –33 –2, –16 –18 –4, –8 –12

26. n2 − 120 = 7n




and


Factors of −120 Sum −1, 120 119 1, −120 −119 −2, 60 58 2, −60 −58 −3, 40 373, −40 −37

−4, 30 264, −30 −26

−5, 24 195, −24 −19

−6, 20 146, −20 −14

−8, 15 78, −15 −7

−12, 10 −212, −10 2

27. d2 + 56 = −18d




and


Factors of 56 Sum 1, 56 57 2, 38 40 4, 14 18 7, 8 15

28. y2 − 90 = 13y




and


Factors of 90 Sum 1, −90 −89 −1, 90 89 2, −45 −43 −2, 45 43 3, −30 −27

−3, 30 275 ,−18 −13

−5, 18 136, −15 −9

−6, 15 99, −10 −1

−9, 10 1

29. h2 + 48 = 16h




and


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





Factors of –72 Sum 1, –72 –71 –1, 72 71 2, –36 –34 –2, 36 34 3, –24 –21

–3, 24 214, –18 –14

–4, 18 146, –12 –6

–6, 12 68, –9 –1

–8, 9 1



SOLUTION:




Factors of −12 Sum 1, −12 −11 −1, 12 11 2, −6 −4 −2, 6 4 3, −4 −1

−3, 4 1






33. q2 + 11qr + 18r

2

SOLUTION:



Factors of 18 Sum 1, 18 19 2, 9 11 3, 6 9

34. x2 − 14xy − 51y2

SOLUTION:



Factors of −51 Sum −1, 51 50 −3, 17 14 −17, 3 −14 −51, 1 −50

35. x2 − 6xy + 5y

2

SOLUTION:




36. a2 + 10ab − 39b2

SOLUTION:



Factors of −39 Sum 1, −39 −38 3, −13 −10 −3, 13 10 −39, 1 −38





Factors of −525 Sum 1, −525 −524 −1, 525 524 5, −105 −100 −5, 105 100 7, −75 −68

−7, 75 6815, −35 −20

−15, 35 2021, −25 −4

−21, 25 4


38. A = x2 + 24x − 81

SOLUTION:






Factors of − 81 Sum 1,−81 −80 −1, 81 80 3, −27 −24 −3, 27 24 9,−9 0

39. A = x2 + 13x − 90

SOLUTION:






Factors of − 90 Sum 1, −90 −89 −1, 90 89 2, −45 −43 −2, 45 43 3, −30 −27

−3, 30 275, −18 −13

−5, 18 136, −15 −9

−6, 15 99, −10 −1

−9, 10 1









Factors of − 16 Sum 1, −16 −15 −1, 16 15 2, −8 −6 −2, 8 6 4, −4 0


42. x2 + kx − 19

SOLUTION:




43. x2 + kx + 14

SOLUTION:




−1, −14 −15 −2, −7 −9

44. x2 − 8x + k , k > 0

SOLUTION:





−1, −7 7 −2, −6 12 −3, −5 15 −4, −4 16

45. x2 − 5x + k , k > 0

SOLUTION:





−1, −4 4 −2, −3 6


SOLUTION:




SOLUTION:




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


SOLUTION:

The trinomial is written in x2 + bx + c =

the graphs coincide. therefore, the trinomial has been ... · therefore, the length and the width...

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