Lesson Objectives

- Demonstrate an understanding of how to multiply two binomials using FOIL
- Demonstrate an understanding of how to factor out the GCF from a polynomial
- Learn how to factor a trinomial into the product of two binomials
- Learn how to determine if a polynomial is prime
- Learn how to factor a trinomial with two variables

## Factoring Trinomials when a = 1

When we see a polynomial with exactly three terms, we refer to the polynomial as a "trinomial". In this lesson, we will learn how to
factor a trinomial into the product of two binomials. We will be working with trinomials of the form:

ax

When a = 1, the trinomial can be rewritten as:

x

Factoring a trinomial of this form is the easiest scenario to deal with. Before we look at the procedure, let's think about how we multiply two binomials using FOIL.

Let's suppose we had:

(x + 5)(x + 7)

F » x • x = x

O » x • 7 = 7x

I » 5 • x = 5x

L » 5 • 7 = 35

We combine like terms to obtain our product. Notice how the O and I steps are what we can combine:

x

x

(x + 5)(x + 7) = x

Now suppose we had to reverse the procedure.

x

The x

Since the coefficient on x

x

The 12x or middle term comes from the O + I step.

The 35 or last term comes from the L step.

The same two integers will need to sum to the coefficient of the middle term (12) and yield a product of the last term (35).

We think about two integers whose sum is 12 and whose product is 35. We can think about the different factors of 35:

1,5,7,35

Of course, we have negative possibilities, but we can exclude these here since everything is positive.

We can see that 5 and 7 are the integers we need.

5 + 7 = 12

5 • 7 = 35

Therefore, we use 5 and 7 to fill in our blanks.

x

Note, the order does not matter, we can switch things up if we would like:

x### Factoring Trinomials of the form: ax

Example 1: Factor each

x

Step 1) Factor out the GCF

In this case, the GCF is 1.

Step 2) The first position of each binomial is given:

(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is 9, and c is 20.

We want two integers whose sum is 9 and whose product is 20.

Factors of 20:

1, 2, 4, 5, 10, 20

4, and 5 are the required integers.

4 + 5 = 9

4 • 5 = 20

Step 4) Use the integers to fill in the blanks

x

Example 2: Factor each

6x

Step 1) Factor out the GCF

6(x

Step 2) The first position of each binomial is given:

6(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is 14 and c is 48. We are working inside of the parentheses since we factored out a 6.

We want two integers whose sum is 14 and whose product is 48.

Factors of 48:

1, 2, 3, 4, 6, 8, 12, 16, 24, 48

6, and 8 are the required integers.

6 + 8 = 14

6 • 8 = 48

Step 4) Use the integers to fill in the blanks

6x

Example 3: Factor each

4x

Step 1) Factor out the GCF

4(x

Step 2) The first position of each binomial is given:

4(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is -6 and c is -16. We are working inside of the parentheses since we factored out a 4.

We want two integers whose sum is (-6) and whose product is (-16).

Since we want a negative product, we know this can only come from opposite signs. One integer will be positive and the other negative. Let's think about the positive factors of 16 and then think about making the signs work:

Factors of 16:

1, 2, 4, 8, 16

-8, and +2 are the required integers.

-8 + 2 = -6

-8 • 2 = -16

Step 4) Use the integers to fill in the blanks

4x

Example 4: Factor each

x

Step 1) Factor out the GCF

In this case, the GCF is 1.

Step 2) The first position of each binomial is given:

(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is -5, and c is -40.

We want two integers whose sum is -5 and whose product is -40.

Since we want a negative product, we know this can only come from opposite signs. One integer will be positive and the other negative. Let's think about the positive factors of 40 and then think about making the signs work:

Factors of 40:

1, 2, 4, 5, 8, 10, 20, 40

There are no such integers that yield a sum of -5 and a product of -40. This means our polynomial is prime.

Example 5: Factor each

x

Step 1) Factor out the GCF

In this case, the GCF is 1.

Step 2) The first position of each binomial is given:

(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is 10y, and c is 21y

(x + __y)(x + __y)

If we think about this, it should make sense. In the FOIL process, the F will not involve a y. The O and I steps will yield a variable part of "xy", and the L step will yield a variable part of "y

Now let's look for two integers whose sum is 10 and whose product is 21.

Factors of 21:

1, 3, 7, 21

3, and 7 are the required integers.

3 + 7 = 10

3 • 7 = 21

If we were to put the y back in:

3y + 7y = 10y

3y • 7y = 21y

Step 4) Use the integers to fill in the blanks

x

ax

^{2}+ bx + cWhen a = 1, the trinomial can be rewritten as:

x

^{2}+ bx + cFactoring a trinomial of this form is the easiest scenario to deal with. Before we look at the procedure, let's think about how we multiply two binomials using FOIL.

Let's suppose we had:

(x + 5)(x + 7)

F » x • x = x

^{2}O » x • 7 = 7x

I » 5 • x = 5x

L » 5 • 7 = 35

We combine like terms to obtain our product. Notice how the O and I steps are what we can combine:

x

^{2}+ 7x + 5x + 35x

^{2}+ 12x + 35(x + 5)(x + 7) = x

^{2}+ 12x + 35Now suppose we had to reverse the procedure.

x

^{2}+ 12x + 35 = (_ + _)(_ + _)The x

^{2}or first term comes from the F step.Since the coefficient on x

^{2}is 1, we know the first position of each binomial is an x.x

^{2}+ 12x + 35 = (x + _)(x + _)The 12x or middle term comes from the O + I step.

The 35 or last term comes from the L step.

The same two integers will need to sum to the coefficient of the middle term (12) and yield a product of the last term (35).

We think about two integers whose sum is 12 and whose product is 35. We can think about the different factors of 35:

1,5,7,35

Of course, we have negative possibilities, but we can exclude these here since everything is positive.

We can see that 5 and 7 are the integers we need.

5 + 7 = 12

5 • 7 = 35

Therefore, we use 5 and 7 to fill in our blanks.

x

^{2}+ 12x + 35 = (x + 5)(x + 7)Note, the order does not matter, we can switch things up if we would like:

x

^{2}+ 12x + 35 = (x + 7)(x + 5)### Factoring Trinomials of the form: ax^{2} + bx + c, a = 1

- When the GCF is not 1, factor out the GCF
- The first position of each binomial is a given (x + _)(x + _)
- Find two integers whose sum is b (the coefficient of the middle term), and whose product is c (constant term)
- Use the integers to fill in the blanks

Example 1: Factor each

x

^{2}+ 9x + 20Step 1) Factor out the GCF

In this case, the GCF is 1.

Step 2) The first position of each binomial is given:

(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is 9, and c is 20.

We want two integers whose sum is 9 and whose product is 20.

Factors of 20:

1, 2, 4, 5, 10, 20

4, and 5 are the required integers.

4 + 5 = 9

4 • 5 = 20

Step 4) Use the integers to fill in the blanks

x

^{2}+ 9x + 20 = (x + 4)(x + 5)Example 2: Factor each

6x

^{2}+ 84x + 288Step 1) Factor out the GCF

6(x

^{2}+ 14x + 48)Step 2) The first position of each binomial is given:

6(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is 14 and c is 48. We are working inside of the parentheses since we factored out a 6.

We want two integers whose sum is 14 and whose product is 48.

Factors of 48:

1, 2, 3, 4, 6, 8, 12, 16, 24, 48

6, and 8 are the required integers.

6 + 8 = 14

6 • 8 = 48

Step 4) Use the integers to fill in the blanks

6x

^{2}+ 84x + 288 = 6(x + 6)(x +8)Example 3: Factor each

4x

^{2}- 24x - 64Step 1) Factor out the GCF

4(x

^{2}- 6x - 16)Step 2) The first position of each binomial is given:

4(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is -6 and c is -16. We are working inside of the parentheses since we factored out a 4.

We want two integers whose sum is (-6) and whose product is (-16).

Since we want a negative product, we know this can only come from opposite signs. One integer will be positive and the other negative. Let's think about the positive factors of 16 and then think about making the signs work:

Factors of 16:

1, 2, 4, 8, 16

-8, and +2 are the required integers.

-8 + 2 = -6

-8 • 2 = -16

Step 4) Use the integers to fill in the blanks

4x

^{2}- 24x - 64 = 4(x + 2)(x - 8)### Prime Polynomials

In some cases, we will not be able to find two integers that sum to b and have a product of c. When a polynomial cannot be factored using integers, we say the polynomial is prime. Let's look at an example.Example 4: Factor each

x

^{2}- 5x - 40Step 1) Factor out the GCF

In this case, the GCF is 1.

Step 2) The first position of each binomial is given:

(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is -5, and c is -40.

We want two integers whose sum is -5 and whose product is -40.

Since we want a negative product, we know this can only come from opposite signs. One integer will be positive and the other negative. Let's think about the positive factors of 40 and then think about making the signs work:

Factors of 40:

1, 2, 4, 5, 8, 10, 20, 40

There are no such integers that yield a sum of -5 and a product of -40. This means our polynomial is prime.

### Factoring Trinomials with Two Variables

We may run into a scenario in which we need to factor a trinomial with two variables. Let's look at an example.Example 5: Factor each

x

^{2}+ 10xy + 21y^{2}Step 1) Factor out the GCF

In this case, the GCF is 1.

Step 2) The first position of each binomial is given:

(x + __)(x + __)

Step 3) Find two integers whose sum is b and whose product is c:

In our case, b is 10y, and c is 21y

^{2}. To make things easy, just place a y in the last position of each binomial and work on the number parts. We will rewrite our binomial as:(x + __y)(x + __y)

If we think about this, it should make sense. In the FOIL process, the F will not involve a y. The O and I steps will yield a variable part of "xy", and the L step will yield a variable part of "y

^{2}".Now let's look for two integers whose sum is 10 and whose product is 21.

Factors of 21:

1, 3, 7, 21

3, and 7 are the required integers.

3 + 7 = 10

3 • 7 = 21

If we were to put the y back in:

3y + 7y = 10y

3y • 7y = 21y

^{2}Step 4) Use the integers to fill in the blanks

x

^{2}+ 10xy + 21y^{2}= (x + 3y)(x + 7y)
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