CR.11: Factoring Trinomials

Learning Outcomes

Factor a trinomial with leading coefficient 1.
Factor trinomials by grouping.

Factoring a Trinomial with Leading Coefficient 1

Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial ${x}^{2}+5x+6$ has a GCF of 1, but it can be written as the product of the factors $\left(x+2\right)$ and $\left(x+3\right)$ .

Trinomials of the form ${x}^{2}+bx+c$ can be factored by finding two numbers with a product of $c$ and a sum of $b$ . The trinomial ${x}^{2}+10x+16$ , for example, can be factored using the numbers $2$ and $8$ because the product of these numbers is $16$ and their sum is $10$ . The trinomial can be rewritten as the product of $\left(x+2\right)$ and $\left(x+8\right)$ .

A General Note: Factoring a Trinomial with Leading Coefficient 1

A trinomial of the form ${x}^{2}+bx+c$ can be written in factored form as $\left(x+p\right)\left(x+q\right)$ where $pq=c$ and $p+q=b$ .

Q & A

Can every trinomial be factored as a product of binomials?

No. Some polynomials cannot be factored. These polynomials are said to be prime.

How To: Given a trinomial in the form ${x}^{2}+bx+c$ , factor it

List factors of $c$ .
Find $p$ and $q$ , a pair of factors of $c$ with a sum of $b$ .
Write the factored expression $\left(x+p\right)\left(x+q\right)$ .

Example: Factoring a Trinomial with Leading Coefficient 1

Factor ${x}^{2}+2x - 15$ .

Show Solution

We have a trinomial with leading coefficient $1,b=2$ , and $c=-15$ . We need to find two numbers with a product of $-15$ and a sum of $2$ . In the table, we list factors until we find a pair with the desired sum.

Factors of $-15$	Sum of Factors
$1,-15$	$-14$
$-1,15$	$14$
$3,-5$	$-2$
$-3,5$	$2$

Now that we have identified $p$ and $q$ as $-3$ and $5$ , write the factored form as $\left(x - 3\right)\left(x+5\right)$ .

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that $\left(x - 3\right)\left(x+5\right)={x}^{2}+2x - 15$ .

Q & A

Does the order of the factors matter?

No. Multiplication is commutative, so the order of the factors does not matter.

Try It

Factor ${x}^{2}-7x+6$ .

Show Solution

$\left(x - 6\right)\left(x - 1\right)$

Factoring by Grouping

Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial $2{x}^{2}+5x+3$ can be rewritten as $\left(2x+3\right)\left(x+1\right)$ using this process. We begin by rewriting the original expression as $2{x}^{2}+2x+3x+3$ and then factor each portion of the expression to obtain $2x\left(x+1\right)+3\left(x+1\right)$ . We then pull out the GCF of $\left(x+1\right)$ to find the factored expression.

A General Note: Factoring by Grouping

To factor a trinomial of the form $a{x}^{2}+bx+c$ by grouping, we find two numbers with a product of $ac$ and a sum of $b$ . We use these numbers to divide the $x$ term into the sum of two terms and factor each portion of the expression separately then factor out the GCF of the entire expression.

How To: Given a trinomial in the form $a{x}^{2}+bx+c$ , factor by grouping

List factors of ${a}\cdot{c}$ .
Find $p$ and $q$ , a pair of factors of ${a}\cdot{c}$ with a sum of $b$ .
Rewrite the original expression as $a{x}^{2}+px+qx+c$ .
Pull out the GCF of $a{x}^{2}+px$ .
Pull out the GCF of $qx+c$ .
Factor out the GCF of the expression.

Example: Factoring a Trinomial by Grouping

Factor $5{x}^{2}+7x - 6$ by grouping.

Show Solution

We have a trinomial with $a=5,b=7$ , and $c=-6$ . First, determine $ac=-30$ . We need to find two numbers with a product of $-30$ and a sum of $7$ . In the table, we list factors until we find a pair with the desired sum.

Factors of $-30$	Sum of Factors
$1,-30$	$-29$
$-1,30$	$29$
$2,-15$	$-13$
$-2,15$	$13$
$3,-10$	$-7$
$-3,10$	$7$

So $p=-3$ and $q=10$ .

\begin{array}{cc} 5 x^{2} - 3 x + 10 x - 6 & Rewrite the original expression as a x^{2} + p x + q x + c . \\ x (5 x - 3) + 2 (5 x - 3) & Factor out the GCF of each part . \\ (5 x - 3) (x + 2) & Factor out the GCF of the expression . \end{array}

$\begin{array}{cc}5{x}^{2}-3x+10x - 6 \hfill & \text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\hfill \\ x\left(5x - 3\right)+2\left(5x - 3\right)\hfill & \text{Factor out the GCF of each part}.\hfill \\ \left(5x - 3\right)\left(x+2\right)\hfill & \text{Factor out the GCF}\text{ }\text{ of the expression}.\hfill \end{array}$

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that $\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6$ .

Try It

Factor the following.

$2{x}^{2}+9x+9$
$6{x}^{2}+x - 1$

Show Solution

$\left(2x+3\right)\left(x+3\right)$
$\left(3x - 1\right)\left(2x+1\right)$

In the next video we see another example of how to factor a trinomial by grouping.

Chapter CR: Corequisite Review

Learning Outcomes

Factoring a Trinomial with Leading Coefficient 1

A General Note: Factoring a Trinomial with Leading Coefficient 1

Q & A

How To: Given a trinomial in the form x2+bx+cx2+bx+c{x}^{2}+bx+c, factor it

Example: Factoring a Trinomial with Leading Coefficient 1

Analysis of the Solution

Q & A

Try It

Factoring by Grouping

A General Note: Factoring by Grouping

How To: Given a trinomial in the form ax2+bx+cax2+bx+ca{x}^{2}+bx+c, factor by grouping

Example: Factoring a Trinomial by Grouping

Analysis of the Solution

Try It

How To: Given a trinomial in the form ${x}^{2}+bx+c$ , factor it

How To: Given a trinomial in the form $a{x}^{2}+bx+c$ , factor by grouping