Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

6.5.4 Factoring Trinomials Using the “ac” Method and Substitution

6.5.4 • Factoring Trinomials Using the “ac” Method and Substitution

Activity

Let’s learn a new method that always works on any trinomial of the form $a x^{2} + b x + c$ .

How to factor trinomials of the form $a x^{2} + b x + c$ using the “ $a c$ ” method:

Step 1 - Factor any GCF. Don’t forget!

Step 2 - Find the product $a c$ .

Step 3 - Find two numbers, $m$ and $n$ , that

multiply to $a c$ : $m \cdot n = a \cdot c$ .
add to $b$ : $m + n = b$ .

Step 4 - Split the middle term using $m$ and $n$ .

$a x^{2} + m x + n x + c$

Step 5 - Factor by grouping.

Step 6 - Check by multiplying the factors.

Factor each trinomial using the “ $a c$ ” method.

1. $16 x^{2} - 32 x + 12$

Compare your answer:

$4 (2 x - 3) (2 x - 1)$

2. $6 y^{2} + y - 15$

Compare your answer:

$(3 y + 5) (2 y - 3)$

3. $48 z^{3} - 102 z^{2} - 45 z$

Compare your answer:

$3 z (8 z + 3) (2 z - 5)$

There will be times when a trinomial does not seem to be in the form $a x^{2} + b x + c$ .

In some of these cases, we can make a substitution that allows the trinomial to be written in the $a x^{2} + b x + c$ form. This is called factoring by substitution.

It is standard to use $u$ for the substitution. The substitution for $u$ might be a variable with an exponent, such as $x^{2}$ , or it might be a binomial, such as $(x - 3)$ .

Factor each trinomial using the substitution method.

4. $y^{4} - y^{2} - 20$

Compare your answer:

$(y^{2} + 4) (y^{2} - 5)$

5. $(x - 5)^{2} + 6 (x - 5) + 8$

Compare your answer:

$(x - 3) (x - 1)$

6. $x^{4} - 3 x^{2} - 28$

Compare your answer:

$(x^{2} - 7) (x^{2} + 4)$

Self Check

Factor using the “

a c

” method:

6x^2 + 13x + 2

.

$(x - 2)(6x + 1)$
$(x - 2)(6x - 1)$
$(6x + 2)(x + 1)$
$(x + 2)(6x + 1)$

Additional Resources

Factoring Trinomials Using the “ $a c$ ” Method

Another way to factor trinomials of the form $a x^{2} + b x + c$ is the “ $a c$ ” method. (The “ $a c$ ” method is sometimes called the grouping method.) The “ $a c$ ” method is actually an extension of the methods you used in the last section to factor trinomials with a leading coefficient of 1. This method is very structured (that is, step by step), and it always works!

Example 1

Factor using the “ $a c$ ” method: $6 x^{2} + 7 x + 2$ .

Step 1 - Factor any GCF.

Is there a greatest common factor?

$6 x^{2} + 7 x + 2$

No!

Step 2 - Find the product $a c$ .

$a x^{2} + b x + c 6 x^{2} + 7 x + 2$

$6 \cdot 2$

$12$

Step 3 - Find two numbers, $m$ and $n$ , that

multiply to $a c$ : $m \cdot n = a \cdot c$ .
add to $b$ : $m + n = b$ .

Find two numbers that multiply to 12 and add to 7. Both factors must be positive.

$3 \cdot 4 = 12$

$3 + 4 = 7$

Step 4 - Split the middle term using $m$ and $n$ .

Rewrite $7 x$ as $3 x + 4 x$ . It would give the same result if we used $4 x + 3 x$ .

Notice that $6 x^{2} + 3 x + 4 x + 2$ is equal to $6 x^{2} + 7 x + 2$ . We just split the middle term to get a more useful form.

Step 5 - Factor by grouping.

$3 x (2 x + 1) + 2 (2 x + 1)$

$(2 x + 1) (3 x + 2)$

Step 6 - Check by multiplying the factors.

$(2 x + 1) (3 x + 2)$

$6 x^{2} + 4 x + 3 x + 2$

$6 x^{2} + 7 x + 2 ✓$

The “ $a c$ ” method is summarized here:

How to factor trinomials of the form $a x^{2} + b x + c$ using the “ $a c$ ” method:

Step 1 - Factor any GCF. Don’t forget!

Step 2 - Find the product $a c$ .

Step 3 - Find two numbers, $m$ and $n$ , that

multiply to $a c$ : $m \cdot n = a \cdot c$ .
add to $b$ : $m + n = b$ .

Step 4 - Split the middle term using $m$ and $n$ .

$a x^{2} + m x + n x + c$

Step 5 - Factor by grouping.

Step 6 - Check by multiplying the factors.

Try it

Factoring Trinomials Using the “ $a c$ ” Method

Factor using the “ $a c$ ” method: $10 y^{2} - 55 y + 70$ .

Here is how to use the “ $a c$ ” method to factor a trinomial.

Step 1 - Is there a greatest common factor?

$10 y^{2} - 55 y + 70$

Yes. The GCF is 5.

Step 2 - Factor it.

$5 (2 y^{2} - 11 y + 14)$

Step 3 - The trinomial inside the parentheses has a leading coefficient that is not $1$ .

$\begin{matrix} a x^{2} + b x + c \\ 5 (2 y^{2} - 11 y + 14) \end{matrix}$

Find the product $a c$ .

$a c = 28$

Step 4 - Find two numbers that multiply to $a c$ and add to $b$ .

$(- 4) (- 7) = 28$

$- 4 + (- 7) = - 11$

Step 5 - Split the middle term.

Step 6 - Factor the trinomial by grouping.

$5 (y (2 y - 7) - 2 (2 y - 7))$

$5 (y - 2) (2 y - 7)$

Step 7 - Check by multiplying all three factors.

$\begin{matrix} 5 (y - 2) (2 y - 7) \\ 5 (2 y^{2} - 7 y - 4 y + 14) \\ 5 (2 y^{2} - 11 y + 14) \\ 10 y^{2} - 55 y + 70 &checkmark; \end{matrix}$

Factoring Trinomials Using Substitution

Sometimes a trinomial does not appear to be in the $a x^{2} + b x + c$ form. However, we can often make a thoughtful substitution that will allow us to make it fit the $a x^{2} + b x + c$ form. This is called factoring by substitution. It is standard to use $u$ for the substitution.

In the form $a x^{2} + b x + c$ , the middle term has a variable, $x$ , and its square, $x^{2}$ , is the variable part of the first term. Look for this relationship as you try to find a substitution.

Example 2

Factor by substitution: $x^{4} - 4 x^{2} - 5$ .

The variable part of the middle term is $x^{2}$ , and its square, $x^{4}$ , is the variable part of the first term. (We know $(x^{2})^{2} = x^{4})$ . If we let $u = x^{2}$ , we can put our trinomial in the $a x^{2} + b x + c$ form we need to factor it.

Step 1 - Rewrite the trinomial to prepare for the substitution.

$(x^{2})^{2} - 4 (x^{2}) - 5$

Step 2 - Let $u = x^{2}$ and substitute.

$u^{2} - 4 u - 5$

Step 3 - Factor the trinomial.

$(u + 1) (u - 5)$

Step 4 - Replace $u$ with $x^{2}$ .

$(x^{2} + 1) (x^{2} - 5)$

Step 5 - Check:

$\begin{matrix} (x^{2} + 1) (x^{2} - 5) \\ x^{4} - 5 x^{2} + x^{2} - 5 \\ x^{4} - 4 x^{2} - 5 &checkmark; \end{matrix}$

Try it

Try It: Factoring Trinomials Using Substitution

Factor by substitution: $(x - 2)^{2} + 7 (x - 2) + 12$ .

In this case, the expression to be substituted is not a monomial. Let $u = x - 2$ .

Here is how to factor using substitution.

The binomial in the middle term, $(x - 2)$ , is squared in the first term. If we let $u = x - 2$ and substitute, our trinomial will be in $a x^{2} + b x + c$ form.

Step 1 - Rewrite the trinomial to prepare for the substitution.

$(x - 2)^{2} + 7 (x - 2) + 12$

Step 2 - Let $u = x - 2$ and substitute.

$u^{2} + 7 u + 12$

Step 3 - Factor the trinomial.

$(u + 3) (u + 4)$

Step 4 - Replace $u$ with $x - 2$ .

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

Step 5 - Simplify inside the parentheses.

$(x + 1) (x + 2)$

This could also be factored by first multiplying out the $(x - 2)^{2}$ and the $7 (x - 2)$ , combining like terms, and then factoring. Most students prefer the substitution method.

6.5.4 Factoring Trinomials Using the “ac” Method and Substitution