1.5 Factoring polynomials (Page 2/6)

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Factor $x (b^{2} - a) + 6 (b^{2} - a)$ by pulling out the GCF.

$(b^{2} - a) (x + 6)$

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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 $x^{2} + 5 x + 6$ has a GCF of 1, but it can be written as the product of the factors $(x + 2)$ and $(x + 3) .$

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

A General Note

Factoring a trinomial with leading coefficient 1

A trinomial of the form $x^{2} + b x + c$ can be written in factored form as $(x + p) (x + q)$ where $p q = c$ 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 $x^{2} + b x + c,$ 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 $(x + p) (x + q) .$

Factoring a trinomial with leading coefficient 1

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

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 [link] , 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 $(x - 3) (x + 5) .$

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

$(x −6) (x −1)$

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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 $2 x^{2} + 5 x + 3$ can be rewritten as $(2 x + 3) (x + 1)$ using this process. We begin by rewriting the original expression as $2 x^{2} + 2 x + 3 x + 3$ and then factor each portion of the expression to obtain $2 x (x + 1) + 3 (x + 1) .$ We then pull out the GCF of $(x + 1)$ to find the factored expression.

A General Note

Factor by grouping

To factor a trinomial in the form $a x^{2} + b x + c$ by grouping, we find two numbers with a product of $a c$ 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 $a x^{2} + b x + c,$ factor by grouping.

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

Factoring a trinomial by grouping

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

We have a trinomial with $a = 5, b = 7,$ and $c = −6.$ First, determine $a c = −30.$ We need to find two numbers with a product of $−30$ and a sum of $7.$ In [link] , 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{matrix} 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{matrix}

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Source: OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3

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