Factoring Trinomials Reverse FOIL Lesson

In the last lesson, we learned how to factor a trinomial of the form: ax² + bx + c, a ≠ 1, using the ac method. In this lesson, we will explore an alternative approach that involves reversing the FOIL procedure. Let's once again think about how to find the product of two binomials using FOIL.
(2x - 5)(x + 7)
F » 2x • x = 2x²
O » 2x • 7 = 14x
I » -5 • x = -5x
L » -5 • 7 = -35
When we combine like terms we get:
2x² + 9x - 35
To reverse the FOIL process, we first think about the F in FOIL.
Since 2, the coefficient of x² is a prime number, our job is a bit easier. To get 2x², we multiply 2x • x:
(2x + __)(x + __)
Now we need to find the missing two integers. We must think about factors of the constant term (-35), that will give us the correct middle term (9x).
-35:
-1 • 35 or 1 • -35
5 • -7 or -5 • 7
At this point, we only need to work out the O and I steps, they must sum to (+9x):
(2x + __)(x + __)
O » 2x • __
I » __ • x
We will start by working with -1 and 35, we have two scenarios.
(2x - 1)(x + 35)
(2x + 35)(x - 1)

O	I	O + I
2x • 35 = 70x	-1 • x = -x	69x
2x • -1 = -2x	35 • x = 35x	33x

Neither produces the correct middle term of 9x. Let's now try +1 and -35, we have two scenarios.
(2x + 1)(x - 35)
(2x - 35)(x + 1)

O	I	O + I
2x • -35 = -70x	1 • x = x	-69x
2x • 1 = 2x	-35 • x = -35x	-33x

Neither produces the correct middle term of 9x. Let's now try +5 and -7, we have two scenarios.
(2x + 5)(x - 7)
(2x - 7)(x + 5)

O	I	O + I
2x • -7 = -14x	5 • x = 5x	-9x
2x • 5 = 10x	-7 • x = -7x	3x

We can see in the top entry that we obtain -9x. Since we want +9x, we just need to change the signs. We will see that our integers are +7 and -5. Let's now try our final possibility of -5 and 7, we have two scenarios:
(2x - 5)(x + 7)
(2x + 7)(x - 5)

O	I	O + I
2x • 7 = 14x	-5 • x = -5x	9x
2x • -5 = -10x	7 • x = 7x	-3x

From the top entry in our table, we can see we get the correct middle term of 9x. This means we can factor our trinomial as:
2x² + 9x - 35 = (2x - 5)(x + 7)
Let's look at an example.
Example 1: Factor each using reverse FOIL.
5x² + 46x + 9
Since 5, the coefficient of x² is a prime number, our job is a bit easier. To get 5x², we multiply 5x • x:
(5x + __)(x + __)
Now we need to find the missing two integers. We must think about the factors of the constant term 9, that will give us the correct middle term (46x).
9:
1 • 9
3 • 3
We don't need to think about negative factors since all signs are positive in our example.
We will start by working with 1 and 9, we have two scenarios:
(5x + 1)(x + 9)
(5x + 9)(x + 1)

O	I	O + I
5x • 9 = 45x	1 • x = 1x	46x
5x • 1 = 5x	9 • x = 9x	14x

From the top entry in our table, we can see we get the correct middle term of 46x. This means we can factor our trinomial as:
5x² + 46x + 9 = (5x + 1)(x + 9)
A more difficult scenario occurs when the coefficient of the squared variable is not prime. When this occurs, we have more possibilities to consider. Let's look at an example.
Example 2: Factor each using reverse FOIL.
20x² + 45x - 140
First and foremost, we can factor out a common factor of 5:
5(4x² + 9x - 28)
We have factored out a 5 and will work on the trinomial inside of the parentheses.
Let's factor:
4x² + 9x - 28
Since 4 is not a prime number, we have to consider more than one possibility.
4x² = 4x • x
4x² = 2x • 2x
We have two scenarios:
(4x + __)(x + __)
(2x + __)(2x + __)
Our final term is -28, to make life easier, let's just think about the factors of +28 and play with the signs:
1 • 28
2 • 14
4 • 7
We will begin with 1 and 28:
(4x + 28)(x + 1)
Since our trinomial 4x² + 9x - 28 does not have a common factor other than 1, the binomials will not either. Notice how 4x and 28 have a common factor of 4. We can quickly eliminate this as a possibility.
(4x + 1)(x + 28)

O	I
4x • 28 = 112x	1 • x = 1x

There's no way to get a middle term of 9x, no matter how we play with the signs.
(2x + 28)(x + 1)
Not possible since 2x and 28 have a common factor of 2.
(2x + 1)(x + 28)

O	I
2x • 28 = 56x	1 • x = 1x

Again, no matter how we play with the signs, we can't get a middle term of 9x.
Let's move on and think about 2 and 14.
The scenario of:
(2x + 14)(2x + 2)
can be eliminated, we have a binomial with 2x and 14 and also 2x and 2, each has a common factor of 2.
(4x + 2)(x + 14)
(4x + 14)(x + 2)
can be eliminated, we have a binomial with 4x and 2, which has a common factor of 2, and a binomial with 4x and 14, which also has a common factor of 2.
Let's move on and think about 4 and 7.
The scenario of:
(2x + 4)(2x + 7)
can be eliminated, we have a binomial with 2x and 4, which has a common factor of 2.
(4x + 4)(x + 7)
can also be eliminated, 4x and 4 have a common factor of 4.
Our final possibility is:
(4x + 7)(x + 4)

O	I
4x • 4 = 16x	7 • x = 7x

If we change the sign of 7 to negative, we will get the correct middle term of 9x.
(4x - 7)(x + 4)

O	I	O + I
4x • 4 = 16x	-7 • x = -7x	9x

Let's not forget that we factored out a 5 at the beginning of the problem. Let's state our final answer as:
20x² + 45x - 140 = 5(4x - 7)(x + 4)

Factoring Trinomials with Two Variables Using Reverse FOIL

We will often come across problems where we need to factor a trinomial with two variables. We can solve these problems using the same techniques we use when there is a single variable. Let's look at an example.
Example 3: Factor each using reverse FOIL.
9x² + 13xy + 4y²
We will end up with y at the end of each binomial:
(__x + __y)(__x + __y)
9 is not a prime number, we can find 9 as:
9 • 1 or 3 • 3
We will have two scenarios:
(9x + __y)(x + __y)
(3x + __y)(3x + __y)
4, the final term is found as:
4 • 1 or 2 • 2
Since all terms are positive, we only need to think about positive numbers.
Let's begin by thinking about 4 and 1:
(9x + 1y)(x + 4y)
(9x + 4y)(x + 1y)
(3x + 1y)(3x + 4y)

O	I	O + I
9x • 4y = 36xy	1y • x = 1xy	37xy
9x • 1y = 9xy	4y • x = 4xy	13xy
3x • 4y = 12xy	1y • 3x = 3xy	15xy

The second entry gives us the correct middle term of 13xy. We have found our factorization as:
9x² + 13xy + 4y² = (9x + 4y)(x + y)

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