Just like in arithmetic, where it is sometimes useful to represent a number in factored form (for example, $12$ as $2\cdot 6\text{ or }3\cdot 4$) in algebra it can be useful to represent a polynomial in factored form. One way to do this is by finding the greatest common factor of all the terms. Remember that you can multiply a polynomial by a monomial as follows:

$\begin{array}{ccc}\hfill 2\left(x + 7\right)&\text{factors}\hfill \\ \hfill 2\cdot x + 2\cdot 7\hfill \\ \hfill 2x + 14&\text{product}\hfill \end{array}$

Here, we will start with a product, like $2x+14$, and end with its factors, $2\left(x+7\right)$. To do this we apply the Distributive Property “in reverse”.

The form on the left is used to multiply. The form on the right is used to factor.

So how do we use the Distributive Property to factor a polynomial? We find the GCF of all the terms and write the polynomial as a product!

example

Factor: $2x+14$

Solution

Step 1: Find the GCF of all the terms of the polynomial.	Find the GCF of $2x$ and $14$.
Step 2: Rewrite each term as a product using the GCF.	Rewrite $2x$ and $14$ as products of their GCF, $2$. $2x=2\cdot x$ $14=2\cdot 7$	$2x+14$ $\color{red}{2}\cdot x+\color{red}{2}\cdot7$
Step 3: Use the Distributive Property ‘in reverse’ to factor the expression.		$2\left(x+7\right)$
Step 4: Check by multiplying the factors.		Check: $2(x+7)$ $2\cdot{x}+2\cdot{7}$ $2x+14\quad\checkmark$

Notice that in the example, we used the word factor as both a noun and a verb:

$\begin{array}{cccc}\text{Noun}\hfill & & & 7\text{ is a factor of }14\hfill \\ \text{Verb}\hfill & & & \text{factor }2\text{ from }2x+14\hfill \end{array}$

example

Factor: $3a+3$

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Solution

	$3a+3$
Rewrite each term as a product using the GCF.	$\color{red}{3}\cdot a+\color{red}{3}\cdot1$
Use the Distributive Property ‘in reverse’ to factor the GCF.	$3(a+1)$
Check by multiplying the factors to get the original polynomial.
$3(a+1)$ $3\cdot{a}+3\cdot{1}$ $3a+3\quad\checkmark$

The expressions in the next example have several factors in common. Remember to write the GCF as the product of all the common factors.

example

Factor: $12x - 60$

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Solution

	$12x-60$
Rewrite each term as a product using the GCF.	$\color{red}{12}\cdot x-\color{red}{12}\cdot 5$
Factor the GCF.	$12(x-5)$
Check by multiplying the factors.
$12(x-5)$ $12\cdot{x}-12\cdot{5}$ $12x-60\quad\checkmark$

Watch the following video to see more examples of factoring the GCF from a binomial.

Now we’ll factor the greatest common factor from a trinomial. We start by finding the GCF of all three terms.

example

Factor: $3{y}^{2}+6y+9$

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Solution

	$3y^2+6y+9$
Rewrite each term as a product using the GCF.	$\color{red}{3}\cdot{y}^{2}+\color{red}{3}\cdot 2y+\color{red}{3}\cdot 3$
Factor the GCF.	$3(y^{2}+2y+3)$
Check by multiplying.
$3(y^{2}+2y+3)$ $3\cdot{y^2}+3\cdot{2y}+3\cdot{3}$ $3y^{2}+6y+9\quad\checkmark$

In the next example, we factor a variable from a binomial.

example

Factor: $6{x}^{2}+5x$

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Solution

	$6{x}^{2}+5x$
Find the GCF of $6{x}^{2}$ and $5x$ and the math that goes with it.
Rewrite each term as a product.	$\color{red}{x}\cdot{6x}+\color{red}{x}\cdot{5}$
Factor the GCF.	$x\left(6x+5\right)$
Check by multiplying.
$x\left(6x+5\right)$ $x\cdot 6x+x\cdot 5$ $6{x}^{2}+5x\quad\checkmark$

When there are several common factors, as we’ll see in the next two examples, good organization and neat work helps!

example

Factor: $4{x}^{3}-20{x}^{2}$

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Solution

		$4x^3-20x^2$
Rewrite each term.		$\color{red}{4{x}^{2}}\cdot x - \color{red}{4{x}^{2}}\cdot 5$
Factor the GCF.		$4x^2(x-5)$
Check.	$4x^2(x-5)$ $4x^2\cdot{x}-4x^2\cdot{5}$ $4x^3-20x^2\quad\checkmark$

example

Factor: $21{y}^{2}+35y$

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Solution

Find the GCF of $21{y}^{2}$ and $35y$
		$21y^2+35y$
Rewrite each term.		$\color{red}{7y}\cdot 3y + \color{red}{7y}\cdot 5$
Factor the GCF.		$7y(3y+5)$

When the leading coefficient, the coefficient of the first term, is negative, we factor the negative out as part of the GCF.

example

Factor: $-9y - 27$

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Solution

When the leading coefficient is negative, the GCF will be negative. Ignoring the signs of the terms, we first find the GCF of $9y$ and $27$ is $9$.
Since the expression $−9y−27$ has a negative leading coefficient, we use $−9$ as the GCF.
	$-9y - 27$
Rewrite each term using the GCF.	$\color{red}{-9}\cdot y + \color{red}{(-9)}\cdot 3$
Factor the GCF.	$-9\left(y+3\right)$
Check. $-9(y+3)$ $-9\cdot{y}+(-9)\cdot{3}$ $-9y-27\quad\checkmark$

Pay close attention to the signs of the terms in the next example.

example

Factor: $-4{a}^{2}+16a$

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Solution

The leading coefficient is negative, so the GCF will be negative.
Since the leading coefficient is negative, the GCF is negative, $−4a$.
	$-4{a}^{2}+16a$
Rewrite each term.	$\color{red}{-4a}\cdot{a}-\color{red}{(-4a)}\cdot{4}$
Factor the GCF.	$-4a\left(a - 4\right)$
Check on your own by multiplying.