Factoring Trinomials AC Method Lesson

In the last lesson, we learned how to factor a trinomial of the form:
ax² + bx + c, a = 1
This means we can write our trinomial as:
x² + bx + c
When a, the coefficient of x² is 1, it makes our job much easier. The first entry of each binomial is a given.
x² + 8x + 15
We can factor the above trinomial as:
x² + 8x + 15 = (x + __)(x + __)
We know the first entry of each binomial is an x. This is from the F step in FOIL. We know that x² comes from x times x. As we work through FOIL, the O + I steps combine to give us the middle term. The L step gives us the final term. To find the last entry of each binomial, we need to find two integers that sum to b, the coefficient of the middle term (8), and multiply to give us c, the constant term (15). We can think about the factors of 15:
1,3,5,15
3, and 5 are the desired integers.
3 + 5 = 8
3 • 5 = 15
Once we find the desired integers, we can use the integers to fill in the blanks.
x² + 8x + 15 = (x + __)(x + __)
x² + 8x + 15 = (x + 3)(x + 5)

How to Factor a Trinomial when the Leading Coefficient is not 1

At some point, we will need to factor the more difficult scenario. This occurs when the coefficient of x² is not 1. This means our first entry of each binomial is not a given and we need to fill in all of the blanks. To factor a trinomial of the form:
ax² + bx + c, a ≠ 1

• Factor out the GCF, when the GCF is not 1
• Find two integers whose product is ac and whose sum is b
• Use the two integers to rewrite the middle term. This will give us a four-term polynomial
• Factor the four-term polynomial using the grouping method

Let's look at a few examples.
Example 1: Factor each.
8x² + 14x - 49
Step 1) Factor out the GCF.
In this case, our GCF is 1.
Step 2) Find two integers whose product is ac and whose sum is b.
a is the coefficient of the squared term (8), while c is the constant (-49).
8 • -49 = -392
b is the coefficient of the middle term (14)
We are looking for a product of -392, and a sum of 14. Since we want a negative product, we know we need a positive and a negative since opposite signs give us a negative product. We also know the absolute value of the positive number is larger than the absolute value of the negative number since the sum is positive. Let's think about the ways we can find a product of +392 and work out the signs.
1 • 392
2 • 196
4 • 98
7 • 56
8 • 49
14 • 28
If we think about 14 and 28, we can make the pair work if we change the sign of 14. Our desired integers are (-14) and (+28).
-14 • 28 = -392
-14 + 28 = 14
Step 3) Use the two integers to rewrite the middle term.
8x² + 14x - 49
8x² + 28x - 14x - 49
Notice how 28x - 14x combines to give us 14x. We now have a four-term polynomial.
Step 4) Factor using grouping
8x² + 28x - 14x - 49
(8x² + 28x) + (-14x - 49)
4x(2x + 7) - 7(2x + 7)
(4x - 7)(2x + 7)
Example 2: Factor each.
12x² - 38x - 14
Step 1) Factor out the GCF.
In this case, our GCF is 2.
2(6x² - 19x - 7)
Step 2) Find two integers whose product is ac and whose sum is b.
a is the coefficient of the squared term (6), while c is the constant (-7).
6 • -7 = -42
b is the coefficient of the middle term (-19)
We are looking for a product of -42, and a sum of -19. Since we want a negative product, we know we need a positive and a negative since opposite signs give us a negative product. We also know the absolute value of the positive number is smaller than the absolute value of the negative number since the sum is negative. Let's think about the ways we can find a product of +42 and work out the signs.
1 • 42
2 • 21
3 • 14
6 • 7
If we think about 2 and 21, we can make the pair work if we change the sign of 21. Our desired integers are (-21) and (+2).
-21 • 2 = -42
-21 + 2 = -19
Step 3) Use the two integers to rewrite the middle term.
12x² - 38x - 14
2(6x² - 19x - 7)
2(6x² - 21x + 2x - 7)
Notice how -21x + 2x combines to give us -19x. We now have a four-term polynomial.
Step 4) Factor using grouping
2(6x² - 21x + 2x - 7)
2((6x² - 21x) + (2x - 7))
2(3x(2x - 7) + 1(2x - 7))
2(3x + 1)(2x - 7)
Example 3: Factor each.
24x³ - 140x² + 200x
Step 1) Factor out the GCF.
In this case, our GCF is 4x.
4x(6x² - 35x + 50)
Step 2) Find two integers whose product is ac and whose sum is b.
a is the coefficient of the squared term (6), while c is the constant (50).
6 • 50 = 300
b is the coefficient of the middle term (-35)
We are looking for a product of +300, and a sum of -35. Since we want a positive product and a negative sum, we know we need two negative numbers. Let's think about the ways we can find a product of +300 and work out the signs.
1 • 300
2 • 150
3 • 100
4 • 75
5 • 60
6 • 50
10 • 30
12 • 25
15 • 20
If we think about 15 and 20, we can make the pair work if we change the sign of both. Our desired integers are (-15) and (-20).
-15 • -20 = 300
-15 + -20 = -35
Step 3) Use the two integers to rewrite the middle term.
24x³ - 140x² + 200x
4x(6x² - 35x + 50)
4x(6x² - 20x - 15x + 50)
Notice how -20x + (-15x) combines to give us -35x. We now have a four-term polynomial.
Step 4) Factor using grouping
4x(6x² - 20x - 15x + 50)
4x((6x² - 20x) + (-15x + 50))
4x(2x(3x - 10) - 5(3x - 10))
4x(2x - 5)(3x - 10)

Factoring Trinomials with Two Variables using the AC Method

In some cases, we will see trinomials where the leading coefficient is not 1 and two variables are involved. These problems can be solved using the same technique. Let's look at an example.
Example 4: Factor each.
4x² - 3xy - 7y²
We can see that we have two variables involved, x and y. Let's work through our regular steps.
Step 1) Factor out the GCF.
In this case, our GCF is 1.
4x² - 3xy - 7y²
Step 2) Find two integers whose product is ac and whose sum is b.
a is the coefficient of the first squared term (4), while c is the final term (-7y²). In this case, -7y² is not really a constant, but we follow the same thought process. -7y² occurs as the final term and is produced by the L step in FOIL.
4 • -7y² = -28y²
b is the coefficient of the middle term (-3y). We think of -3y as what is multiplying the variable x.
We are looking for a product of -28y², and a sum of -3y. Since we want a negative product and a negative sum, we know we need two integers with opposite signs. Let's think about the ways we can find a product of +28 and work out the signs. Leave y out of this for now. We know that y • y = y².
1 • 28
2 • 14
4 • 7
If we think about 4 and 7, we can make the pair work if we change the sign of 7. Our desired integers are (-7) and (+4).
-7y • 4y = -28y²
-7y + 4y = -3y
Step 3) Use the two integers to rewrite the middle term.
4x² - 7xy + 4xy - 7y²
Notice how -7xy + 4xy combines to give us -3xy. We now have a four-term polynomial.
Step 4) Factor using grouping
4x² - 7xy + 4xy - 7y²
4x² + 4xy - 7xy - 7y²
(4x² + 4xy) + (-7xy - 7y²)
4x(x + y) - 7y(x + y)
(4x - 7y)(x + y)

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