Factoring Trinomials II

CHAPTER 6. FACTORING
378
Report the results on your homework as follows.
y
y = x2 + 10x − 24
10
−15 −12
2
5
x
−60
Hence, the solutions of x2 + 10x − 24 = 0 are x = −12 and x = 2. Note
how these agree with the algebraic solution.
6.4
Factoring Trinomials II
1. We proceed as follows:
i) Compare 6x2 + 13x − 5 with ax2 + bx + c and identify a = 6, b = 13, and
c = −5. Note that the leading coefficient is a = 6, so this case is not a
“drop in place” situation.
ii) Calculate ac. Note that ac = (6)(−5), so ac = −30.
iii) List all integer pairs whose product is ac = −30.
1, −30
2, −15
3, −10
5, −6
Second Edition: 2012-2013
−1, 30
−2, 15
−3, 10
−5, 6
6.4. FACTORING TRINOMIALS II
379
iv) Circle the ordered pair whose sum is b = 13.
1, −30
2, −15
3, −10
5, −6
−1, 30
−2, 15
−3, 10
−5, 6
v) If we try to “drop in place”, (x − 2)(x + 15) = 6x2 + 13x − 5. Right off
the bat, the product of the terms in the “First” position does not equal
6x2 . Instead, we must break up the middle term of 6x2 + 13x − 5 into a
sum of like terms using our circled pair of integers −2 and 15.
6x2 + 13x − 5 = 6x2 − 2x + 15x − 5
Now we factor by grouping. Factor 2x out of the first two terms and 5
out of the second two terms.
= 2x (3x − 1) + 5 (3x − 1)
Now we can factor out (3x − 1).
= (2x + 5)(3x − 1)
3. We proceed as follows:
i) Compare 4x2 − x − 3 with ax2 + bx + c and identify a = 4, b = −1, and
c = −3. Note that the leading coefficient is a = 4, so this case is not a
“drop in place” situation.
ii) Calculate ac. Note that ac = (4)(−3), so ac = −12.
iii) List all integer pairs whose product is ac = −12.
1, −12
2, −6
3, −4
−1, 12
−2, 6
−3, 4
iv) Circle the ordered pair whose sum is b = −1.
1, −12
2, −6
3, −4
−1, 12
−2, 6
−3, 4
v) If we try to “drop in place”, (x + 3)(x − 4) = 4x2 − x − 3. Right off the
bat, the product of the terms in the “First” position does not equal 4x2 .
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CHAPTER 6. FACTORING
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Instead, we must break up the middle term of 4x2 − x − 3 into a sum of
like terms using our circled pair of integers 3 and −4.
4x2 − x − 3 = 4x2 + 3x − 4x − 3
Now we factor by grouping. Factor x out of the first two terms and −1
out of the second two terms.
= x (4x + 3) − 1 (4x + 3)
Now we can factor out (4x + 3).
= (x − 1)(4x + 3)
5. We proceed as follows:
i) Compare 3x2 + 19x + 28 with ax2 + bx + c and identify a = 3, b = 19,
and c = 28. Note that the leading coefficient is a = 3, so this case is not
a “drop in place” situation.
ii) Calculate ac. Note that ac = (3)(28), so ac = 84.
iii) List all integer pairs whose product is ac = 84.
1, 84
2, 42
3, 28
4, 21
6, 14
7, 12
−1, −84
−2, −42
−3, −28
−4, −21
−6, −14
−7, −12
iv) Circle the ordered pair whose sum is b = 19.
1, 84
2, 42
3, 28
4, 21
6, 14
7, 12
−1, −84
−2, −42
−3, −28
−4, −21
−6, −14
−7, −12
v) If we try to “drop in place”, (x + 7)(x + 12) = 3x2 + 19x + 28. Right off
the bat, the product of the terms in the “First” position does not equal
3x2 . Instead, we must break up the middle term of 3x2 + 19x + 28 into a
sum of like terms using our circled pair of integers 7 and 12.
3x2 + 19x + 28 = 3x2 + 7x + 12x + 28
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6.4. FACTORING TRINOMIALS II
381
Now we factor by grouping. Factor x out of the first two terms and 4 out
of the second two terms.
= x (3x + 7) + 4 (3x + 7)
Now we can factor out (3x + 7).
= (x + 4)(3x + 7)
7. Compare 12x2 − 23x + 5 with ax2 + bx + c and note that a = 12, b = −23,
and c = 5. Calculate ac = (12)(5), so ac = 60. Start listing the integer pairs
whose product is ac = 60, but be mindful that you need an integer pair whose
sum is b = −23. Note how we ceased listing ordered pairs the moment we
found the pair we needed.
1, 60
2, 30
3, 20
−1, −60
−2, −30
−3, −20
If we try to “drop in place”, (x − 3)(x − 20) = 12x2 − 23x + 5. Right off the bat,
the product of the terms in the “First” position does not equal 12x2 . Instead,
we must break up the middle term of 12x2 − 23x + 5 into a sum of like terms
using our circled pair of integers −3 and −20.
12x2 − 23x + 5 = 12x2 − 3x − 20x + 5
Now we factor by grouping. Factor 3x out of the first two terms and −5 out
of the second two terms.
= 3x (4x − 1) − 5 (4x − 1)
Now we can factor out (4x − 1).
= (3x − 5)(4x − 1)
9. Compare 6x2 + 17x + 7 with ax2 + bx + c and note that a = 6, b = 17, and
c = 7. Calculate ac = (6)(7), so ac = 42. Start listing the integer pairs whose
product is ac = 42, but be mindful that you need an integer pair whose sum
is b = 17. Note how we ceased listing ordered pairs the moment we found the
pair we needed.
1, 42
−1, −42
2, 21
−2, −21
3, 14
If we try to “drop in place”, (x + 3)(x + 14) = 6x2 + 17x + 7. Right off the bat,
the product of the terms in the “First” position does not equal 6x2 . Instead,
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CHAPTER 6. FACTORING
382
we must break up the middle term of 6x2 + 17x + 7 into a sum of like terms
using our circled pair of integers 3 and 14.
6x2 + 17x + 7 = 6x2 + 14x + 3x + 7
Now we factor by grouping. Factor 2x out of the first two terms and 1 out of
the second two terms.
= 2x (3x + 7) + 1 (3x + 7)
Now we can factor out (3x + 7).
= (2x + 1)(3x + 7)
11. Compare 3x2 + 4x − 32 with ax2 + bx + c and note that a = 3, b = 4,
and c = −32. Calculate ac = (3)(−32), so ac = −96. Start listing the integer
pairs whose product is ac = −96, but be mindful that you need an integer pair
whose sum is b = 4. Note how we ceased listing ordered pairs the moment we
found the pair we needed.
1, −96
2, −48
3, −32
4, −24
6, −16
8, −12
−1, 96
−2, 48
−3, 32
−4, 24
−6, 16
−8, 12
If we try to “drop in place”, (x − 8)(x + 12) = 3x2 + 4x − 32. Right off the bat,
the product of the terms in the “First” position does not equal 3x2 . Instead,
we must break up the middle term of 3x2 + 4x − 32 into a sum of like terms
using our circled pair of integers −8 and 12.
3x2 + 4x − 32 = 3x2 − 8x + 12x − 32
Now we factor by grouping. Factor x out of the first two terms and 4 out of
the second two terms.
= x (3x − 8) + 4 (3x − 8)
Now we can factor out (3x − 8).
= (x + 4)(3x − 8)
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6.4. FACTORING TRINOMIALS II
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13. Compare 3x2 + 28x + 9 with ax2 + bx + c and note that a = 3, b = 28,
and c = 9. Calculate ac = (3)(9), so ac = 27. We need an integer pair whose
product is ac = 27 and whose sum is b = 28. The integer pair 1 and 27 comes
to mind.
If we try to “drop in place”, (x + 1)(x + 27) = 3x2 + 28x + 9. Right off
the bat, the product of the terms in the “First” position does not equal 3x2 .
Instead, we must break up the middle term of 3x2 + 28x + 9 into a sum of like
terms using the integer pair 1 and 27.
3x2 + 28x + 9 = 3x2 + x + 27x + 9
Now we factor by grouping. Factor x out of the first two terms and 9 out of
the second two terms.
= x (3x + 1) + 9 (3x + 1)
Now we can factor out (3x + 1).
= (x + 9)(3x + 1)
15. Compare 4x2 − 21x + 5 with ax2 + bx + c and note that a = 4, b = −21,
and c = 5. Calculate ac = (4)(5), so ac = 20. We need an integer pair whose
product is ac = 20 and whose sum is b = −21. The integer pair −1 and −20
comes to mind.
If we try to “drop in place”, (x − 1)(x − 20) = 4x2 − 21x + 5. Right off
the bat, the product of the terms in the “First” position does not equal 4x2 .
Instead, we must break up the middle term of 4x2 − 21x + 5 into a sum of like
terms using the integer pair −1 and −20.
4x2 − 21x + 5 = 4x2 − x − 20x + 5
Now we factor by grouping. Factor x out of the first two terms and −5 out of
the second two terms.
= x (4x − 1) − 5 (4x − 1)
Now we can factor out (4x − 1).
= (x − 5)(4x − 1)
17. Compare 6x2 − 11x − 7 with ax2 + bx + c and note that a = 6, b = −11,
and c = −7. Calculate ac = (6)(−7), so ac = −42. We need an integer pair
whose product is ac = −42 and whose sum is b = −11. The integer pair 3 and
−14 comes to mind.
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CHAPTER 6. FACTORING
384
If we try to “drop in place”, (x + 3)(x − 14) = 6x2 − 11x − 7. Right off
the bat, the product of the terms in the “First” position does not equal 6x2 .
Instead, we must break up the middle term of 6x2 − 11x − 7 into a sum of like
terms using the integer pair 3 and −14.
6x2 − 11x − 7 = 6x2 + 3x − 14x − 7
Now we factor by grouping. Factor 3x out of the first two terms and −7 out
of the second two terms.
= 3x (2x + 1) − 7 (2x + 1)
Now we can factor out (2x + 1).
= (3x − 7)(2x + 1)
19. Note that the GCF of 16x5 , −36x4 , and 14x3 is 2x3 . Factor out this GCF.
16x5 − 36x4 + 14x3 = 2x3 · 8x2 − 2x3 · 18x + 2x3 · 7
= 2x3 (8x2 − 18x + 7)
Next, compare 8x2 − 18x + 7 with ax2 + bx + c and note that a = 8, b = −18,
and c = 7. Start listing the integer pairs whose product is ac = (8)(7), or
ac = 56, but be mindful that you need an integer pair whose sum is b = −18.
1, 56
2, 28
4, 14
−1, −56
−2, −28
−4, −14
Break up the middle term of 8x2 − 18x + 7 into a sum of like terms using our
circled pair of integers −4 and −14.
2x3 (8x2 − 18x + 7) = 2x3 (8x2 − 14x − 4x + 7)
Now we factor by grouping. Factor 2x out of the first two terms and −1 out
of the second two terms.
= 2x3 2x (4x − 7) − 1 (4x − 7)
Now we can factor out (4x − 7).
= 2x3 (2x − 1)(4x − 7)
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6.4. FACTORING TRINOMIALS II
385
21. Note that the GCF of 36x4 , −75x3 , and 21x2 is 3x2 . Factor out this GCF.
36x4 − 75x3 + 21x2 = 3x2 · 12x2 − 3x2 · 25x + 3x2 · 7
= 3x2 (12x2 − 25x + 7)
Next, compare 12x2 − 25x + 7 with ax2 + bx + c and note that a = 12, b = −25,
and c = 7. Start listing the integer pairs whose product is ac = (12)(7), or
ac = 84, but be mindful that you need an integer pair whose sum is b = −25.
−1, −84
−2, −42
−3, −28
−4, −21
1, 84
2, 42
3, 28
4, 21
Break up the middle term of 12x2 − 25x + 7 into a sum of like terms using our
circled pair of integers −4 and −21.
3x2 (12x2 − 25x + 7) = 3x2 (12x2 − 21x − 4x + 7)
Now we factor by grouping. Factor 3x out of the first two terms and −1 out
of the second two terms.
= 3x2 3x (4x − 7) − 1 (4x − 7)
Now we can factor out (4x − 7).
= 3x2 (3x − 1)(4x − 7)
23. Note that the GCF of 6x4 , −33x3 , and 42x2 is 3x2 . Factor out this GCF.
6x4 − 33x3 + 42x2 = 3x2 · 2x2 − 3x2 · 11x + 3x2 · 14
= 3x2 (2x2 − 11x + 14)
Next, compare 2x2 − 11x + 14 with ax2 + bx + c and note that a = 2, b = −11,
and c = 14. Start listing the integer pairs whose product is ac = (2)(14), or
ac = 28, but be mindful that you need an integer pair whose sum is b = −11.
1, 28
2, 14
4, 7
−1, −28
−2, −14
−4, −7
Break up the middle term of 2x2 − 11x + 14 into a sum of like terms using our
circled pair of integers −4 and −7.
3x2 (2x2 − 11x + 14) = 3x2 (2x2 − 7x − 4x + 14)
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CHAPTER 6. FACTORING
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Now we factor by grouping. Factor x out of the first two terms and −2 out of
the second two terms.
= 3x2 x (2x − 7) − 2 (2x − 7)
Now we can factor out (2x − 7).
= 3x2 (x − 2)(2x − 7)
25. Note that the GCF of 16x4 , −36x3 , and −36x2 is 4x2 . Factor out this
GCF.
16x4 − 36x3 − 36x2 = 4x2 · 4x2 − 4x2 · 9x − 4x2 · 9
= 4x2 (4x2 − 9x − 9)
Next, compare 4x2 − 9x − 9 with ax2 + bx + c and note that a = 4, b = −9,
and c = −9. Start listing the integer pairs whose product is ac = (4)(−9), or
ac = −36, but be mindful that you need an integer pair whose sum is b = −9.
1, −36
2, −18
3, −12
−1, 36
−2, 18
Break up the middle term of 4x2 − 9x − 9 into a sum of like terms using our
circled pair of integers 3 and −12.
4x2 (4x2 − 9x − 9) = 4x2 (4x2 + 3x − 12x − 9)
Now we factor by grouping. Factor x out of the first two terms and −3 out of
the second two terms.
= 4x2 x (4x + 3) − 3 (4x + 3)
Now we can factor out (4x + 3).
= 4x2 (x − 3)(4x + 3)
27. Because there is a power of x larger than one, the equation is nonlinear.
Make one side zero.
4x2 = −x + 18
2
4x + x = 18
2
4x + x − 18 = 0
Second Edition: 2012-2013
Original equation.
Add x to both sides.
Subtract 18 from both sides.
6.4. FACTORING TRINOMIALS II
387
Compare 4x2 + x − 18 with ax2 + bx + c and note that a = 4, b = 1, and
c = −18. Calculate ac = (4)(−18), so ac = −72. We need an integer pair
whose product is ac = −72 and whose sum is b = 1. The integer pair −8 and
9 comes to mind. Break up the middle term of 4x2 + x − 18 into a sum of like
terms using the integer pair −8 and 9.
4x2 + 9x − 8x − 18 = 0
x (4x + 9) − 2 (4x + 9) = 0
x = 9x − 8x
Factor by grouping.
(x − 2)(4x + 9) = 0
Factor out 4x + 9.
We have a product that equals zero. Use the zero product property to complete
the solution.
x−2=0
or
4x + 9 = 0
4x = −9
9
x=−
4
x=2
Thus, the solutions of 4x2 = −x + 18 are x = 2 and x = −9/4.
29. Because there is a power of x larger than one, the equation is nonlinear.
Make one side zero.
3x2 + 16 = −14x
Original equation.
2
3x + 14x + 16 = 0
Add 14x to both sides.
Compare 3x2 + 14x + 16 with ax2 + bx + c and note that a = 3, b = 14, and
c = 16. Calculate ac = (3)(16), so ac = 48. We need an integer pair whose
product is ac = 48 and whose sum is b = 14. The integer pair 6 and 8 comes
to mind. Break up the middle term of 3x2 + 14x + 16 into a sum of like terms
using the integer pair 6 and 8.
3x2 + 8x + 6x + 16 = 0
x (3x + 8) + 2 (3x + 8) = 0
14x = 8x + 6x
Factor by grouping.
(x + 2)(3x + 8) = 0
Factor out 3x + 8.
We have a product that equals zero. Use the zero product property to complete
the solution.
x+2=0
x = −2
or
3x + 8 = 0
3x = −8
8
x=−
3
Thus, the solutions of 3x2 + 16 = −14x are x = −2 and x = −8/3.
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CHAPTER 6. FACTORING
388
31. Because there is a power of x larger than one, the equation is nonlinear.
Make one side zero.
3x2 + 30 = 23x
Original equation.
2
3x − 23x + 30 = 0
Subtract 23x from both sides.
Compare 3x2 − 23x + 30 with ax2 + bx + c and note that a = 3, b = −23,
and c = 30. Calculate ac = (3)(30), so ac = 90. We need an integer pair whose
product is ac = 90 and whose sum is b = −23. The integer pair −5 and −18
comes to mind. Break up the middle term of 3x2 − 23x + 30 into a sum of like
terms using the integer pair −5 and −18.
3x2 − 5x − 18x + 30 = 0
x (3x − 5) − 6 (3x − 5) = 0
−23x = −5x − 18x
Factor by grouping.
(x − 6)(3x − 5) = 0
Factor out 3x − 5.
We have a product that equals zero. Use the zero product property to complete
the solution.
x−6=0
x=6
3x − 5 = 0
3x = 5
5
x=
3
Thus, the solutions of 3x2 + 30 = 23x are x = 6 and x = 5/3.
or
33. Because there is a power of x larger than one, the equation is nonlinear.
Make one side zero.
−7x − 3 = −6x2
Original equation.
2
Add 6x2 to both sides.
6x − 7x − 3 = 0
Compare 6x2 − 7x − 3 with ax2 + bx + c and note that a = 6, b = −7, and
c = −3. Calculate ac = (6)(−3), so ac = −18. We need an integer pair whose
product is ac = −18 and whose sum is b = −7. The integer pair 2 and −9
comes to mind. Break up the middle term of 6x2 − 7x − 3 into a sum of like
terms using the integer pair 2 and −9.
6x2 − 9x + 2x − 3 = 0
−7x = −9x + 2x
3x (2x − 3) + 1 (2x − 3) = 0
(3x + 1)(2x − 3) = 0
Factor by grouping.
Factor out 2x − 3.
We have a product that equals zero. Use the zero product property to complete
the solution.
3x + 1 = 0
3x = −1
1
x=−
3
Second Edition: 2012-2013
or
2x − 3 = 0
2x = 3
3
x=
2
6.4. FACTORING TRINOMIALS II
389
Thus, the solutions of −7x − 3 = −6x2 are x = −1/3 and x = 3/2.
35. Because there is a power of x larger than one, the equation is nonlinear.
Make one side zero.
26x − 9 = −3x2
Original equation.
2
Add 3x2 to both sides.
3x + 26x − 9 = 0
Compare 3x2 + 26x − 9 with ax2 + bx + c and note that a = 3, b = 26, and
c = −9. Calculate ac = (3)(−9), so ac = −27. We need an integer pair whose
product is ac = −27 and whose sum is b = 26. The integer pair −1 and 27
comes to mind. Break up the middle term of 3x2 + 26x − 9 into a sum of like
terms using the integer pair −1 and 27.
3x2 − x + 27x − 9 = 0
x (3x − 1) + 9 (3x − 1) = 0
26x = −x + 27x
Factor by grouping.
(x + 9)(3x − 1) = 0
Factor out 3x − 1.
We have a product that equals zero. Use the zero product property to complete
the solution.
x+9=0
x = −9
or
3x − 1 = 0
3x = 1
1
x=
3
Thus, the solutions of 26x − 9 = −3x2 are x = −9 and x = 1/3.
37. Because there is a power of x larger than one, the equation is nonlinear.
Make one side zero.
6x2 = −25x + 9
2
6x + 25x = 9
2
6x + 25x − 9 = 0
Original equation.
Add 25x to both sides.
Subtract 9 from both sides.
Compare 6x2 + 25x − 9 with ax2 + bx + c and note that a = 6, b = 25, and
c = −9. Calculate ac = (6)(−9), so ac = −54. We need an integer pair whose
product is ac = −54 and whose sum is b = 25. The integer pair −2 and 27
comes to mind. Break up the middle term of 6x2 + 25x − 9 into a sum of like
terms using the integer pair −2 and 27.
6x2 + 27x − 2x − 9 = 0
3x (2x + 9) − 1 (2x + 9) = 0
(3x − 1)(2x + 9) = 0
25x = 27x − 2x
Factor by grouping.
Factor out 2x + 9.
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CHAPTER 6. FACTORING
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We have a product that equals zero. Use the zero product property to complete
the solution.
3x − 1 = 0
or
2x + 9 = 0
2x = −9
9
x=−
2
3x = 1
1
x=
3
Thus, the solutions of 6x2 = −25x + 9 are x = 1/3 and x = −9/2.
39. Algebraic solution. The equation is nonlinear, so make one side zero.
That’s already done, so use the ac-method to factor. Note that ac = (2)(−5) =
−10 and the integer pair 1 and −10 has product −10 and sum −9, the coefficient
of x. Split the middle term up using this pair.
2x2 − 9x − 5 = 0
2x2 + x − 10x − 5 = 0
Factor by grouping.
x(2x + 1) − 5(2x + 1) = 0
(x − 5)(2x + 1) = 0
Use the zero product property to set both factors equal to zero, then solve the
resulting equations.
x−5=0
or
2x + 1 = 0
x=5
x=−
1
2
Hence, the solutions are x = 5 and x = −1/2.
Calculator solution. Load the left-hand side of the equation 2x2 − 9x− 5 = 0
into Y1 in the Y= menu, then select 6:ZStandard from the ZOOM menu to
produce the following graph.
Both x-intercepts are visible in the viewing window, but we really should
adjust the WINDOW parameters so that the vertex of the parabola is visible
in the viewing window. Adjust the WINDOW parameters as shown, then push
the GRAPH button to produce the accompanying graph.
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6.4. FACTORING TRINOMIALS II
391
Next, use the 2:zero utility from the CALC menu to find the x-intercepts
of the graph.
Report the results on your homework as follows.
y
y = 2x2 − 9x − 5
20
−10
−0.5
5
10
x
−20
Hence, the solutions of 2x2 − 9x − 5 = 0 are x = −0.5 and x = 5. Note
how these agree with the algebraic solution, especially when you note that
−0.5 = −1/2.
41. Algebraic solution. The equation is nonlinear, so make one side zero.
That’s already done, so use the ac-method to factor. Note that ac = (4)(−15) =
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CHAPTER 6. FACTORING
392
−60 and the integer pair 3 and −20 has product −60 and sum −17, the coefficient of x. Split the middle term up using this pair.
4x2 − 17x − 15 = 0
4x2 + 3x − 20x − 15 = 0
Factor by grouping.
x(4x + 3) − 5(4x + 3) = 0
(x − 5)(4x + 3) = 0
Use the zero product property to set both factors equal to zero, then solve the
resulting equations.
x−5=0
or
4x + 3 = 0
x=5
x=−
3
4
Hence, the solutions are x = 5 and x = −3/4.
Calculator solution. Load the left-hand side of the equation 4x2 −17x−15 =
0 into Y1 in the Y= menu, then select 6:ZStandard from the ZOOM menu
to produce the following graph.
Both x-intercepts are visible in the viewing window, but we really should
adjust the WINDOW parameters so that the vertex of the parabola is visible
in the viewing window. Adjust the WINDOW parameters as shown, then push
the GRAPH button to produce the accompanying graph.
Next, use the 2:zero utility from the CALC menu to find the x-intercepts
of the graph.
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6.4. FACTORING TRINOMIALS II
393
Report the results on your homework as follows.
y
y = 4x2 − 17x − 15
50
−10
−0.75
5
10
x
−50
Hence, the solutions of 4x2 − 17x − 15 = 0 are x = −0.75 and x = 5. Note
how these agree with the algebraic solution, especially when you note that
−0.75 = −3/4.
43. Algebraic solution. The equation is nonlinear, so make one side zero.
Subtract 3x2 and 20x from both sides.
2x3 = 3x2 + 20x
2x3 − 3x2 − 20x = 0
Factor out the GCF.
x(2x2 − 3x − 20) = 0
Note that ac = (2)(−20) = −40 and the integer pair 5 and −8 have product
−40 and sum −3, the coefficient of x. Use this pair to break up the middle
term.
x(2x2 + 5x − 8x − 20) = 0
Second Edition: 2012-2013
CHAPTER 6. FACTORING
394
Factor by grouping.
x(x(2x + 5) − 4(2x + 5)) = 0
x(x − 4)(2x + 5) = 0
Use the zero product property to set all three factors equal to zero, then solve
the resulting equations.
x=0
or
x−4=0
or
2x + 5 = 0
x=4
x=−
5
2
Hence, the solutions are x = 0, x = 4, and x = −5/2.
Calculator solution. Load the left-hand side of the equation 2x3 −3x2 −20x =
0 into Y1 in the Y= menu, then select 6:ZStandard from the ZOOM menu
to produce the following graph.
All three x-intercepts are visible in the viewing window, but we really should
adjust the WINDOW parameters so that the turning points of the parabola
are visible in the viewing window. Adjust the WINDOW parameters as shown,
then push the GRAPH button to produce the accompanying graph.
Next, use the 2:zero utility from the CALC menu to find the x-intercepts
of the graph.
Second Edition: 2012-2013
6.4. FACTORING TRINOMIALS II
395
Report the results on your homework as follows.
y
y = 2x3 − 3x2 − 20x
50
−10
−2.5 0
4
10
x
−50
Hence, the solutions of 2x3 = 3x2 + 20x are x = −2.5, x = 0, and x = 4.
Note how these agree with the algebraic solution, especially when you note
that −2.5 = −5/2.
45. Algebraic solution. The equation is nonlinear, so make one side zero.
Subtract 24x from both sides.
10x3 + 34x2 = 24x
10x3 + 34x2 − 24x = 0
Factor out the GCF.
2x(5x2 + 17x − 12) = 0
Second Edition: 2012-2013
CHAPTER 6. FACTORING
396
Note that ac = (5)(−12) = −60 and the integer pair −3 and 20 have product
−60 and sum 17, the coefficient of x. Use this pair to break up the middle
term.
2x(5x2 − 3x + 20x − 12) = 0
Factor by grouping.
2x(x(5x − 3) + 4(5x − 3)) = 0
2x(x + 4)(5x − 3) = 0
Use the zero product property to set all three factors equal to zero, then solve
the resulting equations.
2x = 0
or
x=0
x+4=0
or
x = −4
5x − 3 = 0
3
x=
5
Hence, the solutions are x = 0, x = −4, and x = 3/5.
Calculator solution. Load the left-hand side of the equation 10x3 + 34x2 −
24x = 0 into Y1 in the Y= menu, then select 6:ZStandard from the ZOOM
menu to produce the following graph.
All three x-intercepts are visible in the viewing window, but we really should
adjust the WINDOW parameters so that the turning points of the parabola
are visible in the viewing window. Adjust the WINDOW parameters as shown,
then push the GRAPH button to produce the accompanying graph.
Next, use the 2:zero utility from the CALC menu to find the x-intercepts
of the graph.
Second Edition: 2012-2013