Factor each polynomial, if possible. If the polynomial cannot be

Transcription

8-7 Solving ax^2 + bx + c = 0
Factor each polynomial, if possible. If the polynomial cannot be factored using integers, write prime .
2
1. 3x + 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.
Sum 31
17
30
Factors of 30
1, 30
2, 15
5, 6
The correct factors are 2 and 15.
2
So, 3x + 17x + 10 = (3x + 2) (x +5).
2
2. 2x + 22x + 56
SOLUTION: 2
The GCF of the terms 2x , 22x, and 56 is 2. Factor this first.
2
2
2x +22x + 56 = 2(x + 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
Factors of 28
1, 28
2, 14
4, 7
Sum 29
16
11
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Page 1
2
8-7 Solving
bx=+(3x
c =+02) (x +5).
So, 3x +ax^2
17x ++10
2
2. 2x + 22x + 56
SOLUTION: 2
2
2
2x +22x + 56 = 2(x + 11x + 28) Distributive Property
Factors of 28
1, 28
2, 14
4, 7
Sum 29
16
11
2
So, 2x + 22x + 56= 2(x + 4)(x + 7).
2
3. 5x − 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.
Factors of 20
Sum 1, –20
–19
–1, 20
19
2, –10
–8
–2, 10
8
4, –5
–1
–4, 5
1
There are no factors of 20 with a sum of –3. This trinomial is prime.
2
4. 3x − 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.
Factors of –60
1, –60
–1, 60
Sum –59
59
Page 2
2, –10
–8
–2, 10
8
4, –5
–1
5 + bx + c = 01
8-7 Solving–4,
ax^2
There are no factors of 20 with a sum of –3. This trinomial is prime.
2
4. 3x − 11x − 20
Factors of –60
1, –60
–1, 60
2, –30
–2, 30
3, –20
–3, 20
4, –15
–4, 15
5, –12
–5, 12
6, –10
–6, 10
Sum –59
59
–28
28
–17
17
–11
11
–7
7
–4
4
The correct factors are 4 and –15
.
2
So, 3x − 11x − 20 = (3x +4)(x – 5).
Solve each equation. Confirm your answers using a graphing calculator.
2
5. 2x + 9x + 9 = 0
SOLUTION: Factor the trinomial.
Solve the equation using the Zero Product Property.
(2x + 3)(x + 3) = 0
Page 3
8-7 Solving
ax^2 + bx + c = 0
2
So, 3x − 11x − 20 = (3x +4)(x – 5).
2
5. 2x + 9x + 9 = 0
(2x + 3)(x + 3) = 0
The roots are
and –3.
2
Confirm the roots using a graphing calculator. Let Y1 = 2x + 9x + 9 and Y2 = 0. Use the intersect option from the
CALC menu to find the points intersection.
Thus, the solutions are
and –3.
2
6. 3x + 17x + 20 = 0
(3x + 5)(x + 4) = 0
Page 4
Thus, theax^2
solutions
8-7 Solving
+ bxare
+ c = 0 and –3.
2
6. 3x + 17x + 20 = 0
(3x + 5)(x + 4) = 0
The roots are
and –4 or about –1.667 and –4. 2
Confirm the roots using a graphing calculator. Let Y1 = 3x + 17x + 20 and Y2 = 0. Use the intersect option from
the CALC menu to find the points intersection.
and –4.
2
7. 3x − 10x + 8 = 0
Page 5
8-7 Solving
+ bxare
+ c = 0 and –4.
Thus, theax^2
solutions
2
7. 3x − 10x + 8 = 0
The roots are
and 2 or about 1.333 and 2. 2
Confirm the roots using a graphing calculator. Let Y1 = 3x –10x + 8 and Y2 = 0. Use the intersect option from the
and 2.
2
8. 2x − 17x + 30 = 0
(2x – 5) (x – 6) = 0
Page 6
8-7 Solving
+ bxare
+ c = and 2.
0
Thus, theax^2
solutions
2
8. 2x − 17x + 30 = 0
(2x – 5) (x – 6) = 0
The roots are
and 6. 2
Confirm the roots using a graphing calculator. Let Y1 = 2x –17x + 30 and Y2 = 0. Use the intersect option from
and 6.
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 by
ofCognero
the discus is for the time t = 0.
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a. What is the initial height of the discus?
8-7 Solving
+ bx +seconds
c = 0 does the discus hit the ground?
b. Afterax^2
how many
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
2.5 seconds. Check by substituting 2.5 in for x in the original equation.
or
Therefore, the discus will hit the ground after 2.5 seconds.
2
10. 5x + 34x + 24
Factors of 120
Sum 1, 120
121
2, 60
62
3, 40
43
4, 30
34
5, 24
29
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6, 20
26
8, 15
23
Page 8
Therefore, the discus will hit the ground after 2.5 seconds.
8-7 Solving
ax^2 + bx + c = 0
2
10. 5x + 34x + 24
Factors of 120
1, 120
2, 60
3, 40
4, 30
5, 24
6, 20
8, 15
10, 12
Sum 121
62
43
34
29
26
23
22
2
So, 5x + 34x + 24= (5x + 4)(x + 6).
2
11. 2x + 19x + 24
Factors of 48
1, 48
2, 24
3, 16
4, 12
6, 8
Sum
49
26
19
16
14
Page 9
2
8-7 Solving
bx +(5x
c=
So, 5x +ax^2
34x ++24=
+ 04)(x + 6).
2
11. 2x + 19x + 24
Factors of 48
1, 48
2, 24
3, 16
4, 12
6, 8
Sum
49
26
19
16
14
2
So, 2x + 19x + 24= (2x + 3)(x + 8).
2
12. 4x + 22x + 10
SOLUTION: 2
2
2
4x + 22x + 10 = 2(2x + 11x + 5) Distributive Property
positive. List the factors of 2(5) or 10 and identify the factors with a sum of 11.
Factors of 10
1, 10
2, 5
Sum
11
7
Page 10
8-7 Solving
ax^2 + bx + c = 0
2
So, 2x + 19x + 24= (2x + 3)(x + 8).
2
12. 4x + 22x + 10
SOLUTION: 2
2
2
4x + 22x + 10 = 2(2x + 11x + 5) Distributive Property
positive. List the factors of 2(5) or 10 and identify the factors with a sum of 11.
Factors of 10
1, 10
2, 5
Sum
11
7
2
So, 4x + 22x + 10= 2(2x + 1)(x + 5).
2
13. 4x + 38x + 70
SOLUTION: 2
2
2
4x +38x + 70 = 2(2x + 19x + 35) Distributive Property
Factors of 70
1, 70
2, 35
5, 14
7, 10
Sum
71
37
19
17
Page 11
8-7 Solving
ax^2 + bx + c = 0
2
So, 4x + 22x + 10= 2(2x + 1)(x + 5).
2
13. 4x + 38x + 70
SOLUTION: 2
2
2
Factors of 70
1, 70
2, 35
5, 14
7, 10
Sum
71
37
19
17
2
So, 4x + 38x + 70= 2(2x + 5)(x + 7).
2
14. 2x − 3x − 9
Factors of –18
1, –18
–1, 18
2, –9
–2, 9
3, –6
–3, 6
Sum
–17
17
–7
7
–3
3
The correct factors are 3 and –6.
Page 12
8-7 Solving
ax^2 + bx + c = 0
2
So, 4x + 38x + 70= 2(2x + 5)(x + 7).
2
14. 2x − 3x − 9
Factors of –18
1, –18
–1, 18
2, –9
–2, 9
3, –6
–3, 6
Sum
–17
17
–7
7
–3
3
2
So, 2x − 3x − 9= (2x + 3)(x − 3).
2
15. 4x − 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.
Factors of 40
–1, –40
–2, –20
–4, –10
–5, –8
Sum
–41
–22
–14
–13
The correct factors are –5 and –8.
Page 13
8-7 Solving
ax^2 + bx + c = 0
2
So, 2x − 3x − 9= (2x + 3)(x − 3).
2
15. 4x − 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
Factors of 40
–1, –40
–2, –20
–4, –10
–5, –8
Sum
–41
–22
–14
–13
2
So, 4x − 13x + 10= (4x − 5)(x − 2).
2
16. 2x + 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
Factors of 12
Sum
1, 12
13
2, 6
8
3, 4
7
There are no factors of 12 with a sum of 3.
2
So, 2x + 3x + 6 is prime.
2
17. 5x + 3x + 4
Factors of 20
Sum
1, 20
21
2, 10
12
4, 5
9
2
2
18. 12x + 69x + 45
SOLUTION: Page 14
1, 12
13
2, 6
8
3, 4
7
There
are
no
factors
of
12
8-7 Solving ax^2 + bx + c = 0with a sum of 3.
2
2
17. 5x + 3x + 4
Factors of 20
Sum
1, 20
21
2, 10
12
4, 5
9
2
2
18. 12x + 69x + 45
SOLUTION: 2
2
2
Factors of 60
1, 60
2, 30
3, 20
4, 15
5, 12
6, 10
Sum
61
32
23
19
17
60
2
So, 12x + 69x + 45= 3(4x + 3)(x + 5).
2
19. 4x − 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
Factors of 28
Sum
Page 15
–1,–28
–29
–2, –14
–16
–4, –7
–11
2
So, 12x + 69x + 45= 3(4x + 3)(x + 5).
2
19. 4x − 5x + 7
Factors of 28
Sum
–1,–28
–29
–2, –14
–16
–4, –7
–11
There are no factors of 28 with a sum of –11.
2
So, 4x − 5x + 7 is prime.
2
20. 5x + 23x + 24
Factors of 120
1, 120
2, 60
3, 40
4, 30
5, 24
6, 20
8, 15
10, 12
Sum
121
62
43
34
29
26
23
22
2
So, 5x + 23x + 24= (5x + 8)(x + 3).
2
21. 3x − 8x + 15
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
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–5, –9
–14
2
8-7 Solving
ax^2 + bx + c = 0
2
So, 5x + 23x + 24= (5x + 8)(x + 3).
2
21. 3x − 8x + 15
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
2
So, 3x − 8x + 15 is prime.
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 + h 0, where h is the height in feet,
t is the time in seconds, v is the initial velocity in feet per second, and h 0 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, h 0, is 6 feet.
2
So, the equation h = −16t + 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.
2
23. 2x + 9x − 18 = 0
Page 17
8-7 Solving
+ c2.= Because
0
The rootsax^2
are + bxand
time cannot be negative, the shot put will hit the ground after 2 seconds.
2
23. 2x + 9x − 18 = 0
The roots are
and –6. 2
Confirm the roots using a graphing calculator. Let Y1 = 2x +9x – 18 and Y2 = 0. Use the intersect option from the
and –6.
2
24. 4x + 17x + 15 = 0
SOLUTION: Factor the trinomial then solve the equation using the Zero Product Property.
.
Page 18
Thus, theax^2
solutions
8-7 Solving
+ bxare
+ c = and –6.
0
2
24. 4x + 17x + 15 = 0
SOLUTION: Factor the trinomial then solve the equation using the Zero Product Property.
.
The roots are
and –3.
2
Confirm the roots using a graphing calculator. Let Y1 = 4x +17x + 15 and Y2 = 0. Use the intersect option from
and –3.
2
25. −3x + 26x = 16
SOLUTION: Factor the trinomial, then solve the equation using the Zero Product Property.
Page 19
8-7 Solving
+ bxare
+ c = 0 and –3.
Thus, theax^2
solutions
2
25. −3x + 26x = 16
The roots are
and 8. 2
Confirm the roots using a graphing calculator. Let Y1 = -3x +26x and Y2 = 16. Use the intersect option from the
and 8.
2
26. −2x + 13x = 15
Page 20
8-7 Solving
+ bxare
+ c = and 8.
0
Thus, theax^2
solutions
2
26. −2x + 13x = 15
The roots are
and 5.
2
Confirm the roots using a graphing calculator. Let Y1 = -2x +13x and Y2 = 15. Use the intersect option from the
and 5.
2
27. −3x + 5x = −2
Page 21
8-7 Solving
+ bxare
+ c = and 5.
0
Thus, theax^2
solutions
2
27. −3x + 5x = −2
The roots are
and 2. 2
Confirm the roots using a graphing calculator. Let Y1 = -3x +5x and Y2 = -2. Use the intersect option from the
and 2.
2
28. −4x + 19x = −30
Page 22
and 2.
2
28. −4x + 19x = −30
The roots are
and 6. 2
Confirm the roots using a graphing calculator. Let Y1 = -4x +19x and Y2 = -30. Use the intersect option from the
and 6.
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?
Page 23
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.
8-7 Solving
+ bxare
+ c = 0 and 6.
Thus, theax^2
solutions
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: 2
a. A model for the vertical motion of a projected object is given by h = −16t + vt + h 0, where h is the height in feet,
t is the time in seconds, v is the initial velocity in feet per second, and h 0 is the initial height in feet. The initial velocity
of the basketball, v, is 20 feet per second. The initial height of the basketball, h 0, is 6 feet. The height of the hoop, h,
is 10 feet.
2
So, the equation 10 = −16t + 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.
2
30. DIVING Ben dives from a 36-foot platform. The equation h = −16t + 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.
Page 24
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.
8-7 Solving
ax^2 + bx + c = 0
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.
2
30. DIVING Ben dives from a 36-foot platform. The equation h = −16t + 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: 2
Let x = a number. Then, 6x +11x = 2.
The possible values of x are –2 or
.
2
32. −6x − 23x − 20
SOLUTION: First factor the number –1 out of each term in the polynomial.
Page 25
The possible values of x are –2 or
.
2
32. −6x − 23x − 20
2
Then factor the trinomial 6x + 23x + 20.
In this trinomial, a = 6, b = 23 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be
Factors of 120
1, 120
2, 60
3, 40
4, 30
5, 24
6, 20
8, 15
10, 12
Sum
121
62
43
34
29
26
23
22
2
So, −6x − 23x − 20= −(2x + 5)(3x + 4).
2
33. −4x − 15x − 14
2
2
−4x − 15x − 14 = –1(4x + 15x + 14)
2
ThenManual
factor- Powered
the trinomial
4x +
eSolutions
by Cognero
15x + 14.
Page 26
2
8-7 Solving
c = 0+ 5)(3x + 4).
So, −6x ax^2
− 23x+−bx
20=+ −(2x
2
33. −4x − 15x − 14
2
2
−4x − 15x − 14 = –1(4x + 15x + 14)
2
Then factor the trinomial 4x + 15x + 14.
Factors of 56
1, 56
2, 28
4, 14
7, 8
Sum
57
30
18
115
2
So, −4x − 15x − 14 = −(x + 2)(4x + 7).
2
34. −5x + 18x + 8
2
2
−5x + 18x + 8 = –1(5x – 18x – 8)
2
Then factor the trinomial 5x – 18x – 8.
In this trinomial, a = 5, b = –18 and c = –8, so m + p is negative and mp is negative. Therefore, m and p must have
Factors of –40
–1, 40
eSolutions Manual - Powered
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–2, 20
2, –20
Sum
39
–39
18
–18
Page 27
8-7 Solving
ax^2 + bx + c = 0
2
So, −4x − 15x − 14 = −(x + 2)(4x + 7).
2
34. −5x + 18x + 8
2
2
−5x + 18x + 8 = –1(5x – 18x – 8)
2
Then factor the trinomial 5x – 18x – 8.
Factors of –40
–1, 40
1, –40
–2, 20
2, –20
–4, 10
4, –10
–5, 8
5, –8
Sum
39
–39
18
–18
6
–6
3
–3
The correct factors are –20 and 2
.
2
So, −5x + 18x + 8 = −(x – 4)(5x + 2).
2
35. −6x + 31x − 35
2
2
−6x + 31x − 35 = –1(6x – 31x + 35)
2
Then factor the trinomial 6x – 31x + 35.
In this trinomial, a = 6, b = –31 and c = 35, so m + p is negative and mp is positive. Therefore, m and p must both be
Page 28
of 6(35) or 210 and identify the factors with a sum of –31.
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negative.
the negative
factors
Factors of 210
Sum 8-7 Solving
ax^2 + bx + c = 0
2
So, −5x + 18x + 8 = −(x – 4)(5x + 2).
2
35. −6x + 31x − 35
2
2
−6x + 31x − 35 = –1(6x – 31x + 35)
2
Sum –211
–107
–73
–47
–41
–37
–31
–29
Factors of 210
–1, –210
–2, –105
–3, –70
–5, –42
–6, –35
–7, –30
–10, –21
–14, –15
2
So, −6x + 31x − 35 = −(2x − 7)(3x − 5).
2
36. −4x + 5x − 12
2
2
−4x + 5x − 12 = –1(4x – 5x + 12)
2
Factors of 48
Sum –1, –48
–49
–24 by Cognero
–26
eSolutions Manual - –2,
Powered
Page 29
–3, –16
–19
–4, –12
–16
8-7 Solving
ax^2 + bx + c = 0
2
So, −6x + 31x − 35 = −(2x − 7)(3x − 5).
2
36. −4x + 5x − 12
2
2
−4x + 5x − 12 = –1(4x – 5x + 12)
2
Factors of 48
Sum –1, –48
–49
–2, –24
–26
–3, –16
–19
–4, –12
–16
–6, –8
–14
2
There are no factors of 48 with a sum of –5. So, −4x + 5x − 12 is prime.
2
37. −12x + x + 20
2
2
−12x + x + 20 = –1(12x – x – 20)
2
Then factor the trinomial 12x – x – 20.
Factors of –240
–1, 240
1, –240
–2, 120
2, –120
–3, 80
3, –80
–4, 60
4, –60
–5, 48
5, –48
–6, 40
6, –40
–8, 30
8, –30
–10, 24
10, –24
–12, 20
12, –20
–15, 16
15, –16
eSolutions
Manual - Powered by Cognero
The correct factors are –16 and 15.
Sum
239
–239
118
–118
77
–77
56
–56
43
–43
34
–34
22
–22
14
–14
8
–8
1
–1
Page 30
8, –30
–10, 24
10, –24
–12,
8-7 Solving ax^2
+ 20
bx + c = 0
12, –20
–15, 16
15, –16
–22
14
–14
8
–8
1
–1
2
So, −12x + x + 20 = −(4x + 5)(3x – 4).
38. URBAN PLANNING The city has commissioned the building of a rectangular park. The area of the park can be
2
expressed as 660x + 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
eSolutions
Page 31
dimensions?
d. ANALYTICAL Write the area of the rectangle as the product of two binomials.
e. VERBAL Complete this statement: a 2 − b 2 = … Why is this statement true?
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: a 2 − b 2 = … Why is this statement true?
SOLUTION: a.
2
2
The area of a square is found using the formula A = s . So, the area of the larger square is a and the area of the
2
smaller square is b .
b. To find the area of the remaining region, subtract the area of the smaller square from the area of the larger square.
2
2
So, the area of the remaining region is a – b .
c.
The width is a – b and the length is a + b.
d.
e . The figure with area a 2 − b 2 and the rectangle with area (a − b)(a + b) have the same area, so a 2 − b 2 = (a − b)
(a + b).
2
40. CCSS CRITIQUE Zachary and Samantha are solving 6x − x = 12. Is either of them correct? Explain your
reasoning.
Page 32
2
2
2
2
e . The figure
8-7 Solving
ax^2with
+ bxarea
+ c a= 0− b and the rectangle with area (a − b)(a + b) have the same area, so a − b = (a − b)
(a + b).
2
40. CCSS CRITIQUE Zachary and Samantha are solving 6x − 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.
2
2
41. REASONING A square has an area of 9x + 30xy + 25y square inches. The dimensions are binomials with
positive integer coefficients. What is the perimeter of the square? Explain.
SOLUTION: 2
2
The area of the square equals 9x + 30xy + 25y .
So, the length of each side of the square is (3x + 5y) inches.
The perimeter of the square is (12x + 20y) inches.
2
42. CHALLENGE Find all values of k so that 2x + k x + 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
Page 33
The perimeter of the square is (12x + 20y) inches.
2
42. CHALLENGE Find all values of k so that 2x + k x + 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
2
polynomial of the form ax + bx + c.
SOLUTION: 2
2
Sample answer: If the trinomial factors, then ax + bx + c = ax + mx + px + c where m and n are two integers that
2
have a product equal to ac and a sum equal b. For example, given the trinomial 3x + 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 the
variables 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.
Page 34
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
3 5
a b and
5 2
a b ?
8 7
A a b
B a b
C a b
D a b
2 3
8 3
2 7
SOLUTION: Choice A is the correct answer.
2
47. What is the solution set of x + 2x − 24 = 0?
F {−4, 6}
G {−3, 8}
H {3,
eSolutions
Manual
−8}- Powered by Cognero
J {4, −6}
Page 35
Choice Aax^2
is the+ correct
8-7 Solving
bx + c answer.
=0
2
47. What is the solution set of x + 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.
2
49. x − 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.
eSolutions Manual
- Powered
Page 36
Factors
of by
14Cognero
Sum
–1, –14
–15
–2, –7
–9
SOLUTION: Becauseax^2
x is greater
8-7 Solving
+ bx +than
c = or
0 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.
2
49. x − 9x + 14
Factors of 14
Sum
–1, –14
–15
–2, –7
–9
2
50. n − 8n + 15
Factors of 15
Sum
–1, –15
–16
–3, –5
–8
2
51. x − 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
23
–1, 24
2, –12
–10
10
–2, 12
3, –8
–5
5
–3, 8
4, –6
–2
2
–4, 6
2
52. z + 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 by
36Cognero
Sum
eSolutions Manual
- Powered
Page 37
1, 36
37
2, 18
20
–4, 6
2
2
52. z + 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
2
53. r + 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
39
–1, 40
2, –20
–18
18
–2, 20
4, –10
–6
6
–4, 10
5, –8
–3
3
–5, 8
2
54. v + 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: eSolutions
Page 38
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.
2
57. 10x − 20x = 0
SOLUTION: eSolutions Manual - Powered by Cognero
Page 39
The solutions are –3 and 1.
2
57. 10x − 20x = 0
SOLUTION: The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.
and
2
58. 8b − 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
2
.
Page 40
2
58. 8b − 12b = 0
in for b in the original equation.
and
.
2
59. 15a = 60a
and
eSolutions2Manual - Powered by Cognero
60. 33x = −22x
SOLUTION: Page 41
8-7 Solving
ax^2 are
+ bx
+ c = .0
The solutions
0 and
2
59. 15a = 60a
and
2
60. 33x = −22x
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
Page 42
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.
.
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: and
The solution set is {k|10 < k ≤ 16}.
To graph the solution set, graph k ≤ 16 and graph k > 10. Then find the intersection.
63. d − 4 > 3 or d − 4 ≤ 1
SOLUTION: or
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.
64. 3 < 2x − 3 < 15
SOLUTION: and
Page 43
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.
64. 3 < 2x − 3 < 15
SOLUTION: and
The solution set is {x|3 < x < 9}.
To graph the solution set, graph x < 9 and graph x > 3. Then find the intersection.
65. 3t − 7 ≥ 5 and 2t + 6 ≤ 12
SOLUTION: and
The solution set is empty. A number cannot be both greater than or equal to 4 and less than or equal to 3.
66. h − 10 < −21 or h + 3 < 2
SOLUTION: or
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.
67. 4 < 2y − 2 < 10
SOLUTION: and
The solution set is {y|3 < y < 6}.
To graph the solution set, graph y < 6 and graph y > 3. Then find the intersection.
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.
Page 44
The solution
8-7 Solving
ax^2set+ isbx{y|3
+ c<=y 0< 6}.
To graph the solution set, graph y < 6 and graph y > 3. Then find the intersection.
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
Page 45
73. 121
SOLUTION: 8-7 Solving ax^2 + bx + c = 0
74. 100
Page 46

Factor each polynomial, if possible. If the polynomial cannot be

Transcription

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