Chapter 1: Factoring and Quadratic Equations
Section 1.4: Factoring Trinomials in the Form ax2 + bx + c where a ≠ 1
#1 – 18: Rewrite as a polynomial with 4 terms (if possible) then factor by grouping and check
your answer, state if a polynomial is prime.
1) 5x2 + 11x + 6
First multiply the 5*6 = 30
Factors of 30
Choice to break up middle
term
What middle column
simplifies to
1 * 30
1x + 30x
31x
2 * 15
2x + 15x
17x
3 * 10
3x + 10x
13x
5*6
5x + 6x
11x
= 5x2 + 5x + 6x + 6
= (5x2 + 5x) + (6x + 6)
= 5x(x+1) + 6(x+1)
Solution: (x+1)(5x+6)
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Chapter 1: Factoring and Quadratic Equations
3) 5x2 – 11x + 6
First multiply the 5*6 = 30
Factors of 30
Choice to break up middle
term
What middle column
simplifies to
-1 * -30
-1x + -30x
-31x
-2 * -15
-2x + -15x
-17x
-3 * -10
-3x + -10x
-13x
-5 * -6
-5x + -6x
-11x
Factors of 30
Choice to break up middle
term
What middle column
simplifies to
1 * 12
1x + 12x
13x
2*6
2x + 6x
8x
3*4
3x + 4x
7x
= 5x2 - 5x - 6x + 6
= (5x2 - 5x) + (-6x + 6)
= 5x(x-1) + -6(x-1)
= 5x(x -1) – 6(x -1)
Solution: (x-1)(5x-6)
5) 4x2 + 7x + 3
First multiply the 4*3 = 12
= 4x2 + 4x + 3x + 3
= (4x2 + 4x) + (3x + 3)
= 4x(x+1) + 3(x+1)
Solution: (x+1)(4x+3)
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Chapter 1: Factoring and Quadratic Equations
7) 4x2 – 7x + 3
First multiply the 4*3 = 12
Factors of 30
Choice to break up middle
term
What middle column
simplifies to
-1*-12
-1x + - 12x
-13x
-2*-6
-2x + - 6x
-8x
-3*-4
-3x + - 4x
-7x
Factors of 30
Choice to break up middle
term
What middle column
simplifies to
-1*14
-1z + 14z
13z
1*-14
1z + -14z
-13z
-2*7
-2z + 7z
5z
2*-7
2z + -7z
-5z
= 4x2 + -3x + -4x + 3
= x(4x - 3) + (-1)(4x – 3)
= x(4x - 3) - 1(4x – 3)
Solution: (4x – 3)(x – 1)
9) 2z2 + 5z – 7
First multiply the 2*-7= -14
= 2z2 + (-2z) + 7z – 7
= 2z(z-1) + 7(z-1)
Solution: (z-1)(2z+7)
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Chapter 1: Factoring and Quadratic Equations
11) 2z2 – 5z – 7
First multiply the 2*-7= -14
Factors of -14
Choice to break up middle
term
What middle column
simplifies to
-1*14
-1z + 14z
13z
1*-14
1z + -14z
-13z
-2*7
-2z + 7z
5z
2*-7
2z + -7z
-5z
Factors of 42
Choice to break up middle
term
What middle column
simplifies to
1 * 42
1x + 42x
43x
2 * 21
2x + 21x
23x
3 * 14
3x + 14x
17x
6*7
6x + 7x
13x
= 2z2 + 2z + - 7z -7
= 2z(z+1) + -7(z+1)
Solution: (z+1)(2z – 7)
13) 6x2 + 23x + 7
First multiply the 6*7 = 42
= 6x2 + 2x + 21x + 7
= (6x2 + 2x) + (21x + 7)
= 2x(3x+1) + 7(3x+1)
Solution: (3x+1)(2x+7)
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Chapter 1: Factoring and Quadratic Equations
15) 3x2 + 10x + 7
First multiply the 3*7 = 21
Factors of 21
Choice to break up middle
term
What middle column
simplifies to
1 * 21
1x + 21x
23x
3*7
3x + 7x
10x
Factors of 30
Choice to break up middle
term
What middle column
simplifies to
1 * 30
1x + 30x
31x
2 * 15
2x + 15x
17x
3 * 10
3x + 10x
13x
5*6
5x + 6x
11x
3x2 + 3x + 7x + 7
= 3x(x + 1) + 7(x +1)
Solution: (x+1)(3x + 7)
17) 5x2 + 13x + 6
First multiply the 5*6 = 30
= 5x2 + 3x + 10x + 6
= (5x2 + 3x) + (10x + 6)
= x(5x+3) + 2(5x+3)
Solution: (5x+3)(x+2)
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Chapter 1: Factoring and Quadratic Equations
#19-44: Factor, using bottoms up or the guess and check method, state if a polynomial is prime
(notice #19 – 30 are the same as #1-12, and you should get the same answer regardless of the
technique you use to factor.)
19) 5x2 + 11x + 6
First multiply the 5*6 = 30
Rewrite as x2 + 11x + 30 and factor this
Factors of 30
Sum of factors
1 * 30
1+30 = 31
2 * 15
2 + 15 = 17
3 * 10
3 + 10 = 13
5*6
5 + 6 = 11
(x + 5)(x+6)
Finish divide by 5 / reduce / bottoms up
5
6
(𝑥 + 5) (𝑥 + 5)
Solution: (x+1)(5x+6)
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Chapter 1: Factoring and Quadratic Equations
21) 5x2 – 11x + 6
First multiply the 5*6 = 30
Rewrite as x2 - 11x + 30 and factor this
Factors of 30
Sum of factors
-1 * -30
-1+-30 = -31
-2 * -15
-2 + -15 = -17
-3 * -10
-3 + -10 = -13
-5 * -6
-5 + -6 = -11
(x - 5)(x-6)
Finish divide by 5 / reduce / bottoms up
5
6
(𝑥 − 5) (𝑥 − 5)
Solution: (x-1)(5x-6)
23) 4x2 + 7x + 3
First multiply 4*3 = 12 and rewrite the problem
x2 + 7x + 12
Factor this
(x+3)(x+4)
Now divide by 4 / reduce and bottoms up
3
4
(𝑥 + 4) (𝑥 + 4)
Solution: (4x+ 3)(x+1)
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Chapter 1: Factoring and Quadratic Equations
25) 4x2 – 7x + 3
First multiply 4*3 = 12 and rewrite the problem
x2 - 7x + 12
Factor this
(x-3)(x-4)
Now divide by 4 / reduce and bottoms up
3
4
(𝑥 − 4) (𝑥 − 4)
Solution: (4x- 3)(x-1)
27) 2z2 + 5z – 7
First multiply the 2*-7 = -14 and rewrite the problem
z2 + 5z – 14
factor this (z+7)(z-2)
7
2
now divide by 2 / reduce and bottoms up (𝑧 + 2) (𝑧 − 2)
Solution: (2z + 7)(z – 1)
29) 2z2 – 5z – 7
First multiply the 2*-7 = -14 and rewrite the problem
z2 + 5z – 14
factor this (z-7)(z+2)
7
2
now divide by 2 / reduce and bottoms up (𝑧 − 2) (𝑧 + 2)
Solution: (2z - 7)(z + 1)
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Chapter 1: Factoring and Quadratic Equations
31) 3x2 – 11x + 10
First multiply the 3*10 = 30 and rewrite the problem
x2 – 11x + 30
factor this (x – 5)(x – 6)
5
6
now divide by 3 / reduce and bottoms up (𝑥 − 3) (𝑥 − 3)
Solution: (3x – 5)(x – 2)
33) 2b2 – 15b + 7
First multiply 2 * 7 = 14 and rewrite the problem
b2 – 15b + 14 factor this
(b – 14)(b – 1)
Now divide by 2 / reduce and bottoms up
= (𝑏 −
14
2
1
) (𝑏 − 2)
Solution: (b – 7)(2b – 1)
35) 6y2 – 7y – 5
First multiply the 6*-5 = -30 and rewrite the problem
x2 – 7x - 30
factor this (x + 3)(x - 10)
3
now divide by 6 / reduce and bottoms up (𝑥 + 6) (𝑥 −
1
10
6
)
5
(𝑥 + 2) (𝑥 − 3)
Solution: (2x + 1)(3x – 5)
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Chapter 1: Factoring and Quadratic Equations
37) 8a2 + a – 7
First multiply 8* -7 = -56 and rewrite the problem
a2 + a – 56 and factor this
(a+8)(a-7)
Now divide by 8 / reduce and bottoms up
8
7
(𝑎 + 8) (𝑎 − 8)
Solution: (a + 1)(8a – 7)
39) 2x2 – 5x – 7
First multiply 2* -7 = -14 and rewrite the problem
x2 - 5x - 14 and factor this
(x + 2)(x – 7)
Now divide by 2 / reduce and bottoms up
2
7
(𝑥 + 2) (𝑥 − 2)
Solution: (x + 1)(2x – 7)
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Chapter 1: Factoring and Quadratic Equations
41) 3x2 + 5x + 6
First multiply 3*6 = 18 and rewrite the problem
x2 + 5x + 18
my choices for the lasts are 1*9 or 2*9 or 3*6
none of these gives me the 5 that I need so this is prime
Solution: Prime
43) 2x2 + 5x – 8
First multiply 2*-8 = -16 and rewrite the problem
x2 + 5x - 16
This is prime, so the original problem is also prime.
Solution: prime
#45-64: Factor out the GCF and then factor by bottoms up or the guess and check method.
45) 4m2 + 34m – 18
First factor out the GCF of 2
= 2(2m2 + 17m – 9)
Now bottoms up factor the inside of the parenthesis
2(m2 + 17m – 18)
2(m + 18)(m – 1)
2 (𝑚 +
18
1
) (𝑚 − 2)
2
Solution: 2(m+9)(2m-1)
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Chapter 1: Factoring and Quadratic Equations
47) 4z3 – 13z2 + 3z
First factor out the GCF of z
= z(4z2 – 13z + 3)
Now bottoms up factor the inside of the parenthesis
z(z2 – 13z + 12)
z(z – 1)(z – 12)
1
𝑧 (𝑧 − 4) (𝑧 −
12
4
)
Solution: z(4z – 1)(z – 3)
49) 20x3 – 18x2 + 4x
First factor out the GCF of 2x
= 2x(10x2 – 9x + 2)
Now I will bottom up what is left inside the parenthesis.
2x(x2 – 9x + 20)
2x(x – 5)(x – 4)
5
4
2𝑥 (𝑥 − 10) (𝑥 − 10)
1
2
2𝑥 (𝑥 − 2) (𝑥 − 5)
Solution: 2x(2x – 1)(5x – 2)
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Chapter 1: Factoring and Quadratic Equations
51) -16x2 + 44x – 10
First factor out the GCF of -2
= -2(8x2 – 22x + 5)
Now bottoms up factor what’s left in the parenthesis.
-2(x2 – 22x + 40)
-2(x – 20)(x – 2)
−2 (𝑥 −
20
2
) (𝑥 − 8)
8
5
1
−2 (𝑥 − 2) (𝑥 − 4)
Solution: -2(2x – 5)(4x – 1)
53) 18x2 – 21x – 15
First factor out the GCF of 3
= 3(6x2 – 7x – 5)
Now bottoms up factor the inside of the parenthesis
3(x2 – 7x – 30)
3(x+ 3)(x-10)
3 (𝑥 + 6) (𝑥 −
3
10
1
5
6
)
3 (𝑥 + 2) (𝑥 − 3)
Solution: 3(2x+1)(3x-5)
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Chapter 1: Factoring and Quadratic Equations
55) 18x3 – 21x2 – 15x
First factor out the GCF of 3x
= 3x(6x2 – 7x – 5)
Now bottoms up factor the inside of the parenthesis
3x(x2 – 7x – 30)
3x(x+ 3)(x-10)
3
3𝑥 (𝑥 + 6) (𝑥 −
1
10
6
)
5
3𝑥 (𝑥 + 2) (𝑥 − 3)
Solution: 3x(2x+1)(3x-5)
57) -18x2 + 21x + 15
First factor out the GCF of -3
= -3(6x2 – 7x – 5)
Now bottoms up factor the inside of the parenthesis
-3(x2 – 7x – 30)
-3(x+ 3)(x-10)
3
−3 (𝑥 + 6) (𝑥 −
1
10
6
)
5
−3 (𝑥 + 2) (𝑥 − 3)
Solution: -3(2x+1)(3x-5)
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Chapter 1: Factoring and Quadratic Equations
59) 12x2 + 10x + 12
First factor out GCF of 2
= 2(6x2 + 5x + 6)
Now bottoms up what is left inside the parenthesis.
2(x2 + 5x + 36)
the inside of the parenthesis is prime, so I can’t factor more. It would be wrong to write prime
as my answer as I was able to factor out a GCF of 2.
Solution: 2(6x2 + 5x + 6)
61) 3x3 + 5x2 + 6x
First factor out the GCF of x
= x(3x2 + 5x + 6)
Now bottoms up factor what is left
x(x2 + 5x + 18)
the inside of the parenthesis is prime, so I can’t factor more. It would be wrong to write prime
as my answer as I was able to factor out a GCF of x.
Solution: x(3x2 + 5x + 6)
63) 4b3 + 6b2 - 6b
factor out GCF of 2b
= 2b(2b2 + 3b – 3)
now bottoms up factor the inside of the parenthesis
2b(b2 + 3b – 6)
the inside of the parenthesis is prime, so I can’t factor more. It would be wrong to write prime
as my answer as I was able to factor out a GCF of 2b.
Solution: 2b(2b2 + 3b – 3)
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