1. Factoring Trinomials of the form
MTH Algebra
Factoring Trinomials of the form
ax2 + bx + c
a ≠ 1
Chapter 5 Section 4

2. Factoring Trinomials ax2 + bx + c, a ≠ 1
The squared term has a numerical coefficient not equal to 1.
There are two methods
Trial and Error
Factor by Grouping.
Remember factoring is the reverse of multiplication.

3. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Trial and Error
Determine whether there is a factor common to all three terms. If yes, factor it out. (we do not factor out 1 or -1)
Write all pairs of factors of the coefficient of the squared term, a.
Write all pairs of factors of the constant term, c.
Try various combination of these factors until the correct middle term, bx, is found.
NOTE: You may want to select a style when deciding the position of the factors of the coefficient of the squared term (Exp: put the larger value in the first binomial)

4. Trial and Error Factors of 2 Example: 2x2 + 11x + 12
Possible factors
of trinomial
Products of the Outer and Inner Terms
Sums to the b term in the Trinomial
(1)(12)
(2x + 1)(x + 12)
(2x)(12) + (1)(x) = 24x + x = 25x
(2)(6)
(2x + 2)(x + 6)
(2x)(6) + (2)(x) = 12x + 2x = 14x
(3)(4)
(2x + 3)(x + 4)
(2x)(4) + (3)(x) = 8x + 3x = 11x
(12)(1)
(2x + 12)(x + 1)
(2x)(1) + (12)(x) = 2x + 12x = 14x
(6)(2)
(2x + 6)(x + 2)
(2x)(2) + (6)(x) = 4x + 6x = 10x
(4)(3)
(2x + 4)(x + 3)
(2x)(3) + (4)(x) = 6x + 4x = 10x
Since the constant is positive and the x-term is positive both factors are positive.
You can always check with the FOIL method

5. Trial and Error Factors of 7 Example: 7x2 – 11x – 6
Possible factors
of trinomial
Products of the Outer and Inner Terms
Sums to the b term in the Trinomial
(-1)(6)
(7x - 1)(x + 6)
(7x)(6) + (-1)(x) = 42x + -x = 41x
(-2)(3)
(7x - 2)(x + 3)
(7x)(3) + (-2)(x) = 21x + -2x = 19x
(-3)(2)
(7x - 3)(x + 2)
(7x)(2) + (-3)(x) = 14x + -3x = 11x
(-6)(1)
(7x - 6)(x + 1)
(7x)(1) + (-6)(x) = 7x + -6x = 1x
(1)(-6)
(7x + 1)(x - 6)
(7x)(-6) + (1)(x) = -42x + 1x = -41x
(2)(-3)
(7x + 2)(x - 3)
(7x)(-3) + (2)(x) = -21x + 2x = -19x
(3)(-2)
(7x + 3)(x - 2)
(7x)(-2) + (3)(x) = -14x + 3x = -11x
(6)(-1)
(7x + 6)(x - 1)
(7x)(-1) + (6)(x) = -7x + 6x = -1x
Since the last term is negative, one factors is positive and one is negative.
When we change the sign of the constant in the binomial , x-term in trinomial changes)

6. Trial and Error Factors of 6 Example: 6x2 + 31x + 5
(2)(3)
Factors of 5
Possible factors
of trinomial
Products of the Outer and Inner Terms
Sums to the b term in the Trinomial
(1)(5)
(6x + 1)(x + 5)
(6x)(5) + (1)(x) = 30x + 1x = 31x
(5)(1)
(6x + 5)(x + 1)
(6x)(1) + (5)(x) = 6x + 5x = 11x
(2x + 1)(3x + 5)
(2x)(5) + (1)(3x) = 10x + 3x = 13x
(2x + 5)(3x + 1)
(2x)(1) + (5)(3x) = 2x + 15x = 17x
Since the constant is positive and the middle-term is positive both factors are positive.
You can always check with the FOIL method

7. Trial and Error Example: 16x2 - 8x + 1 (16)(1) (4x - 1)(4x - 1) (2)(8)
Factors of 16
Example: x2 - 8x (16)(1)
(4x - 1)(4x - 1) (2)(8)
(4x - 1)2 (4)(4)
Factors of 1
Possible factors
of trinomial
Products of the Outer and Inner Terms
Sums to the b term in the Trinomial
(-1)(-1)
(1x - 1)(x - 1)
(1x)(-1) + (-1)(x) = -1x + -1x = -2x
(2X - 1)(8X - 1)
(2X)(-1) + (-1)(8X) = -2X + -8X = -10X
(4X - 1)(4X - 1)
(4X)(-1) + (-1)(4X) = -4X + -4X = -8X
Since the constant is positive and the middle-term is negative both factors are negative.
You can always check with the FOIL method

8. Trial and Error Factors of 2 Example: 2x2 + 11x + 7 PRIME (2)(1)
Possible factors
of trinomial
Products of the Outer and Inner Terms
Sums to the b term in the Trinomial
(1)(7)
(2x + 1)(x + 7)
(2x)(7) + (1)(x) = 14x + 1x = 15x
(7)(1)
(2x + 7)(x + 1)
(2x)(1) + (7)(x) = 2x + 7x = 9x
Since the constant is positive and the middle-term is positive both factors are positive.
You can always check with the FOIL method

9. Trial and Error Factors of 6 Example: 6a2 + 11ab + 5b2
(6a + 5b)(a + b) (6)(1)
(2)(3)
Factors of 5
Possible factors
of trinomial
Products of the Outer and Inner Terms
Sums to the b term in the Trinomial
(1)(5)
(6a + 1)(a + 5)
(6a)(5) + (1)(a) = 30a + 1a = 31a
(5)(1)
(6a + 5)(a + 1)
(6a)(1) + (5)(a) = 6a + 5a = 11a
(2a + 1)(3a + 5)
(2a)(5) + (1)(3a) = 10a + 3a = 13a
(2a + 5)(3a + 1)
(2a)(1) + (5)(3a) = 2a + 15a = 17a
Since the last term is positive and the middle-term is positive both factors are positive.
You can always check with the FOIL method

10. Trial and Error
Factors of 6
Example: 6x2 – 19xy – 7y2 (6)(1) (3x + y)(2x – 7y) (2)(3)
Factors of 7
Possible factors
of trinomial
Products of the Outer and Inner Terms
Sums to the b term in the Trinomial
(-1)(7)
(6x - 1)(x + 7)
(6x)(7) + (-1)(x) = 42x + -1x = 41x
(-7)(1)
(6x - 7)(x + 1)
(6x)(1) + (-7)(x) = 6x + -7x = -1x
(1)(-7)
(2x + 1)(3x - 7)
(2x)(-7) + (1)(3x) = -14x + 3x = -11x
(7)(-1)
(2x + 7)(3x - 1)
(2x)(-1) + (7)(3x) = -2x + 21x = 19x
(2x - 1)(3x + 7)
(2x)(7) + (-1)(3x) = 14x - 3x = 11x
(2x - 7)(3x + 1)
(2x)(1) + (-7)(3x) = 2x - 21x = -19x
Since the last term is negative one factors is positive and one is negative.

11. Trial and Error
Factors of 3
Example: 9x3 + 15x2 + 6x GCF (3)(1) x(3x2 + 5x + 2x) 3x
3x(3x + 2)(x + 1)
Factors of 2
Possible factors
of trinomial
Products of the Outer and Inner Terms
Sums to the b term in the Trinomial
(1)(2)
(3x + 1)(x + 2)
(3x)(2) + (1)(x) = 6x + 1x = 7x
(2)(1)
(3x + 2)(x + 1)
(3x)(1) + (2)(x) = 3x + 2x = 5x
Since the constant is positive and the middle-term is positive both factors are positive.

12. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping
Determine whether there is a factor common to all three terms. If yes, factor it out.
Find two numbers whose product is equal to the product of a times c, and whose sum is equal to b.
Rewrite the middle term, bx, as the sum or difference of two terms using the numbers found in step 2.
Factor by grouping.

13. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping
Example:
3x2 + 14x + 15
3x2 + 9x + 5x + 15
3x(x + 3) + 5(x + 3)
(3x + 5)(x + 3)
a = 3, b = 14, c = 15
a • c = 3 • 15 = 45
Factors of 45
Sum 14
(1)(45)
=46
(3)(15)
= 18
(5)(9)
5 + 9 = 14
Since a • c is positive and the middle-term is positive both factors are positive.

14. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping
Example:
3x2 - 7x - 6
3x2 - 9x + 2x - 6
3x(x - 3) + 2(x - 3)
(3x + 2)(x - 3)
a = 3, b = -7, c = -6
a • c = 3 • -6 = -18
Factors of -18
Sum -7
(-1)(18)
= 17
(-2)(9)
= 7
(-3)(6)
= 3
(1)(-18)
= -17
(2)(-9)
= -7
(3)(-6)
= -3
Since a • c is negative one factors is positive and one is negative.

15. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping
Example:
6x2 + 31x + 5
6x2 + x + 30x + 5
x(6x + 1) + 5(6x + 1)
(6x + 1)(x + 5)
a = 6, b = 31, c = 5
a • c = 6 • 5 = 30
Factors of 30
Sum 31
(1)(30)
= 31
Since a • c is positive and the middle-term is positive both factors are positive.

16. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping
Example:
64p2 – 16p + 1
64p2 - 8p - 8p + 1
8p(8p - 1) - 1(8p - 1)
(8p - 1)(8p - 1)
(8p – 1)2
a = 64, b = -16, c = 1
a • c = 64 • 1 = 64
Factors of 64
Sum -16
(-1)(-64)
= -65
(-2)(-32)
= -34
(-4)(-16)
= -20
(-8)(-8)
= -16
Since a • c is positive and the middle-term is negative both factors are negative.

17. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping
Example:
3x2 + 20x + 5
3x2 + ____ + 5
PRIME
a = 3, b = 20, c = 5
a • c = =3 • 5 = 15
Factors of 15
Sum 5
(1)(15)
= 16
(3)(5)
3 + 5 = 8
Since a • c is positive and the middle-term is positive both factors would have been positive.

18. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping
Example:
5a2 + 9ab + 4b2
5a2 + 4ab + 5ab + 4b2
5a2 + 5ab + 4ab + 4b2
5a(a + b) + 4b(a + b)
(a + b)(5a + 4b)
a = 5, b = 9, c = 4
a • c = 5 • 4 = 20
Factors of 20
Sum 9
(1)(20)
= 21
(2)(10)
= 12
(4)(5)
4 + 5 = 9
Since a • c is positive and the middle-term is positive both factors are positive.

19. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping
Example:
6x2 – 25xy – 9y2
6x2 + 2xy – 27xy – 9y2
6x2 + 2xy – 27xy + 9y2
2x(3x + y) – 9y(3x + y)
(2x – 9y)(3x + y)
a = 6, b = -25, c = -9
a • c = 6 • -9 = -54
Factors of -54
Sum -25
(-1)(54)
= 53
(-2)(27)
= 25
(1)(-54)
= -53
(2)(-27)
= -25
Since a • c is negative one factors is positive and one is negative.

20. Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping
Example:
6x3 + 3x2 – 45x
GCF = 3x
3x(2x2 + x - 15)
3x(2x2 + 6x - 5x - 15)
3x(2x)(x + 3) - 5(x + 3)
3x(2x – 5)(x + 3)
a = 2, b = 1, c = -15
a • c = 2 • -15 = -30
Factors of -30
Sum 1
(-1)(30)
= 29
(-2)(15)
= 13
(-3)(10)
= 7
(-5)(6)
= 1
Since a • c is negative one factors is positive and one is negative.

21. REMEMBER
Always put the polynomial in standard form before attempting to factor.
For Trial and Error you my want to setup a table listing all the possible factors of a and c, and then use these to form all possible binomial factor pairs for the polynomial.
If the original expression has no common factor, then the two factors will not have any common factors.
Check your results by multiplying.
I prefer Grouping. Which method do you prefer?

22. HOMEWORK 5.4
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