Lesson Objectives

- Demonstrate an understanding of how to factor
- Demonstrate an understanding of how to multiply two binomials using FOIL
- Learn how to factor a trinomial using the AC method
- Learn how to factor a trinomial using the reverse FOIL method

## How to Factor Trinomials when the Leading Coefficient is not 1

In our last lesson, we learned how to factor trinomials of the form:

ax

This means we covered the easy case scenario in which we factored:

x

Having a coefficient of 1 for our x

(x + __)(x + __)

To factor a trinomial of this form, we only needed to figure out two integers that would sum to b (the coefficient of x) and give a product of c (constant). As an example, suppose we want to factor:

x

We could set up our known part as:

(x + __)(x + __)

Now we just need two integers which sum to 5 and give a product of 6. Two such integers would be 2 and 3. We can place the integers in the blanks in any order:

x

Example 1: Factor each

12x

Step 1) Factor out any common factor other than 1 or (-1)

In this case, we can factor out a 3:

3(4x

Now we will only focus on what is inside of the parentheses.

Step 2) Look for two integers whose product is ac and whose sum is b:

a = 4, b = 9, c = 2

a • c » 4 • 2 = 8

We need to find two integers whose sum is 9 and whose product is 8. Let's think about factors of 8:

1, 8

2, 4

1 • 8 = 8

1 + 8 = 9

Our two integers are 1 and 8.

Step 3) Use these two integers to rewrite the middle term so that we have a four-term polynomial.

3(4x

Notice how we changed 9x into (8x + 1x). This is legal since (8x + 1x = 9x).

Step 4) Factor the four-term polynomial using the grouping method:

3(4x

3(4x(x + 2) + 1(x + 2))

3(x + 2)(4x + 1)

Example 2: Factor each

8x

Step 1) Factor out any common factor other than 1 or (-1)

In this case, we can factor out a 2:

2(4x

Now we will only focus on what is inside of the parentheses.

Step 2) Look for two integers whose product is ac and whose sum is b:

a = 4, b = 7, c = -15

a • c » 4 • -15 = -60

We need to find two integers whose sum is 7 and whose product is -60. Let's think about factors of 60 and then work out the signs:

1, 60

2, 30

3, 20

4, 15

5, 12

6, 10

Looking through our factors, think about the difference between each pair. We can then work out the sign afterward. If we subtract 12 - 5, we get 7. We have found our two integers (12, -5):

-5 • 12 = -60

-5 + 12 = 7

Step 3) Use these two integers to rewrite the middle term so that we have a four-term polynomial.

2(4x

Notice how we changed 7x into (12x - 5x). This is legal since (12x - 5x = 7x).

Step 4) Factor the four-term polynomial using the grouping method:

2(4x

2(4x(x + 3) + (-5)(x + 3))

2(x + 3)(4x - 5)

(5x + 1)(x + 5)

F » 5x • x = 5x

O » 5x • 5 = 25x

I » 1 • x = x

L » 1 • 5 = 5

We know that we can combine like terms and obtain:

5x

How could we reverse this process? Let's start with the first term 5x

(5x + __)(x + __)

Now we think about the last term of 5. We know the two blank spots need to multiply together to give us 5. Again, since 5 is a prime number, we know the only possibilities are 5 and 1. We need to check the outer and inner products, these need to sum to 26x:

Trial 1:

(5x + 5)(x + 1)

Outer » 5x

Inner » 5x

O + I = 5x + 5x = 10x

10x ≠ 26x

Since we did not successfully obtain the correct middle term, we try another configuration:

(5x + 1)(x + 5)

Outer » 25x

Inner » 1x

O + I = 25x + 1x = 26x

We have the correct middle term, therefore we have found our correct factorization. Let's take a look at another example.

Example 3: Factor each

7x

Step 1) Let's begin by looking out our first term (7x

(7x + __)(x + __)

Step 2) Let's think about our last term (-18). We will look at the factors of +18 and then play with the signs:

1, 18

2, 9

3, 6

Step 3) Any combination of one positive and one negative would give us a negative product. Let's think about the possibilities:

We want to achieve a middle term of (-15x). This comes from the O + I in the FOIL process. In the outer step, 7x would multiply one factor and in the inner step, 1x would multiply the other. Let's just work with positives for a moment. We begin with the factors 1 and 18:

7 • 1 = 7

7 • 18 = 126

1 • 1 = 1

1 • 18 = 18

There's no way to make the factors 1 and 18 work no matter the signs, let's move on to the next pair of factors, 2 and 9:

7 • 9 = 63

7 • 2 = 14

1 • 9 = 9

2 • 9 = 18

Again, no way to make the factors 2 and 9 work no matter the signs, let's move on to the next pair of factors, 3 and 6:

7 • 3 = 21

7 • 6 = 42

1 • 3 = 3

1 • 6 = 6

The coefficient of the middle term, -15, can come from (-21 + 6). This means we need to multiply (7x • -3) and (1x • 6). We will factor our trinomial as:

7x

ax

^{2}+ bx + c, a = 1This means we covered the easy case scenario in which we factored:

x

^{2}+ bx + cHaving a coefficient of 1 for our x

^{2}part meant that the first term of each binomial was x:(x + __)(x + __)

To factor a trinomial of this form, we only needed to figure out two integers that would sum to b (the coefficient of x) and give a product of c (constant). As an example, suppose we want to factor:

x

^{2}+ 5x + 6We could set up our known part as:

(x + __)(x + __)

Now we just need two integers which sum to 5 and give a product of 6. Two such integers would be 2 and 3. We can place the integers in the blanks in any order:

x

^{2}+ 5x + 6 = (x + 2)(x + 3)### Factoring Trinomials using the AC Method

What can we do when the coefficient on the x^{2}part is not 1? This is the harder scenario and often gives students some grief when starting out. There are many techniques available for this type of problem, but we will only focus on two such solutions. The first is known as factoring using the AC method or grouping. Essentially we are expanding our trinomial into a four-term polynomial and then using the factoring by grouping method we learned a few lessons ago.- Factor out any common factor other than 1 or (-1)
- Look for two integers whose product is ac (a multiplied by c) and whose sum is b
- Use these two integers to rewrite the middle term so that we have a four-term polynomial
- Factor the four-term polynomial using the grouping method

Example 1: Factor each

12x

^{2}+ 27x + 6Step 1) Factor out any common factor other than 1 or (-1)

In this case, we can factor out a 3:

3(4x

^{2}+ 9x + 2)Now we will only focus on what is inside of the parentheses.

Step 2) Look for two integers whose product is ac and whose sum is b:

a = 4, b = 9, c = 2

a • c » 4 • 2 = 8

We need to find two integers whose sum is 9 and whose product is 8. Let's think about factors of 8:

1, 8

2, 4

1 • 8 = 8

1 + 8 = 9

Our two integers are 1 and 8.

Step 3) Use these two integers to rewrite the middle term so that we have a four-term polynomial.

3(4x

^{2}+ 8x + 1x + 2)Notice how we changed 9x into (8x + 1x). This is legal since (8x + 1x = 9x).

Step 4) Factor the four-term polynomial using the grouping method:

3(4x

^{2}+ 8x + 1x + 2)3(4x(x + 2) + 1(x + 2))

3(x + 2)(4x + 1)

Example 2: Factor each

8x

^{2}+ 14x - 30Step 1) Factor out any common factor other than 1 or (-1)

In this case, we can factor out a 2:

2(4x

^{2}+ 7x - 15)Now we will only focus on what is inside of the parentheses.

Step 2) Look for two integers whose product is ac and whose sum is b:

a = 4, b = 7, c = -15

a • c » 4 • -15 = -60

We need to find two integers whose sum is 7 and whose product is -60. Let's think about factors of 60 and then work out the signs:

1, 60

2, 30

3, 20

4, 15

5, 12

6, 10

Looking through our factors, think about the difference between each pair. We can then work out the sign afterward. If we subtract 12 - 5, we get 7. We have found our two integers (12, -5):

-5 • 12 = -60

-5 + 12 = 7

Step 3) Use these two integers to rewrite the middle term so that we have a four-term polynomial.

2(4x

^{2}+ 12x - 5x - 15)Notice how we changed 7x into (12x - 5x). This is legal since (12x - 5x = 7x).

Step 4) Factor the four-term polynomial using the grouping method:

2(4x

^{2}+ 12x - 5x - 15)2(4x(x + 3) + (-5)(x + 3))

2(x + 3)(4x - 5)

### Factoring Trinomials using Reverse FOIL

Alternatively, when we factor a trinomial and the leading coefficient is not 1, we can use the reverse FOIL method. The reverse FOIL method involves undoing the FOIL process through trial and error. Let's first multiply two binomials together using FOIL:(5x + 1)(x + 5)

F » 5x • x = 5x

^{2}O » 5x • 5 = 25x

I » 1 • x = x

L » 1 • 5 = 5

We know that we can combine like terms and obtain:

5x

^{2}+ 26x + 5How could we reverse this process? Let's start with the first term 5x

^{2}. This is the result of the F in FOIL. This means first times first. 5 is a prime number and can only come from 5 times 1. This means we can set up our first terms as:(5x + __)(x + __)

Now we think about the last term of 5. We know the two blank spots need to multiply together to give us 5. Again, since 5 is a prime number, we know the only possibilities are 5 and 1. We need to check the outer and inner products, these need to sum to 26x:

Trial 1:

(5x + 5)(x + 1)

Outer » 5x

Inner » 5x

O + I = 5x + 5x = 10x

10x ≠ 26x

Since we did not successfully obtain the correct middle term, we try another configuration:

(5x + 1)(x + 5)

Outer » 25x

Inner » 1x

O + I = 25x + 1x = 26x

We have the correct middle term, therefore we have found our correct factorization. Let's take a look at another example.

Example 3: Factor each

7x

^{2}- 15x - 18Step 1) Let's begin by looking out our first term (7x

^{2}). 7 is a prime number and can only come from 1 times 7:(7x + __)(x + __)

Step 2) Let's think about our last term (-18). We will look at the factors of +18 and then play with the signs:

1, 18

2, 9

3, 6

Step 3) Any combination of one positive and one negative would give us a negative product. Let's think about the possibilities:

We want to achieve a middle term of (-15x). This comes from the O + I in the FOIL process. In the outer step, 7x would multiply one factor and in the inner step, 1x would multiply the other. Let's just work with positives for a moment. We begin with the factors 1 and 18:

7 • 1 = 7

7 • 18 = 126

1 • 1 = 1

1 • 18 = 18

There's no way to make the factors 1 and 18 work no matter the signs, let's move on to the next pair of factors, 2 and 9:

7 • 9 = 63

7 • 2 = 14

1 • 9 = 9

2 • 9 = 18

Again, no way to make the factors 2 and 9 work no matter the signs, let's move on to the next pair of factors, 3 and 6:

7 • 3 = 21

7 • 6 = 42

1 • 3 = 3

1 • 6 = 6

The coefficient of the middle term, -15, can come from (-21 + 6). This means we need to multiply (7x • -3) and (1x • 6). We will factor our trinomial as:

7x

^{2}- 15x - 18 = (7x + 6)(x - 3)
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