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### Course: Algebra (all content) > Unit 10

Lesson 12: Factoring polynomials by taking common factors- Factoring polynomials by taking a common factor
- Taking common factor from binomial
- Taking common factor from trinomial
- Taking common factor: area model
- Factoring polynomials: common binomial factor
- Factor polynomials: common factor
- Factoring by common factor review
- Factoring polynomials: common factor (old)
- Factoring polynomials: common factor (old)

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# Factoring polynomials by taking a common factor

Learn how to factor a common factor out of a polynomial expression. For example, factor 6x²+10x as 2x(3x+5).

#### What you should be familiar with before this lesson

The $6x$ and $4{x}^{2}$ is $2x$ .

**GCF**(greatest common factor) of two or more monomials is the product of all their common prime factors. For example, the GCF ofIf this is new to you, you'll want to check out our greatest common factors of monomials article.

#### What you will learn in this lesson

In this lesson, you will learn how to factor out common factors from polynomials.

## The distributive property: $a(b+c)=ab+ac$

To understand how to factor out common factors, we must understand the

**distributive property**.For example, we can use the distributive property to find the product of $3{x}^{2}$ and $4x+3$ as shown below:

Notice how each term in the binomial was multiplied by a common factor of ${3{x}^{2}}$ .

However, because the distributive property is an equality, the reverse of this process is also true!

If we start with $3{x}^{2}(4x)+3{x}^{2}(3)$ , we can use the distributive property to ${3{x}^{2}}$ and obtain $3{x}^{2}(4x+3)$ .

**factor out**The resulting expression is in

*factored form*because it is written as a*product*of two polynomials, whereas the original expression is a two-termed sum.### Check your understanding

## Factoring out the greatest common factor (GCF)

To factor the GCF out of a polynomial, we do the following:

- Find the GCF of all the terms in the polynomial.
- Express each term as a product of the GCF and another factor.
- Use the distributive property to factor out the GCF.

Let's factor the GCF out of $2{x}^{3}-6{x}^{2}$ .

**Step 1: Find the GCF**

$2{x}^{3}={2}\cdot {x}\cdot {x}\cdot x$ $6{x}^{2}={2}\cdot 3\cdot {x}\cdot {x}$

So the GCF of $2{x}^{3}-6{x}^{2}$ is ${2}\cdot {x}\cdot {x}={2{x}^{2}}$ .

**Step 2: Express each term as a product of**${2{x}^{2}}$ and another factor.

$2{x}^{3}=({2{x}^{2}})(x)$ $6{x}^{2}=({2{x}^{2}})(3)$

So the polynomial can be written as $2{x}^{3}-6{x}^{2}=({2{x}^{2}})(x)-({2{x}^{2}})(3)$ .

**Step 3: Factor out the GCF**

Now we can apply the distributive property to factor out ${2{x}^{2}}$ .

**Verifying our result**

We can check our factorization by multiplying $2{x}^{2}$ back into the polynomial.

Since this is the same as the original polynomial, our factorization is correct!

### Check your understanding

### Can we be more efficient?

If you feel comfortable with the process of factoring out the GCF, you can use a faster method:

*Once we know the GCF, the factored form is simply the product of that GCF and the sum of the terms in the original polynomial divided by the GCF.*

See, for example, how we use this fast method to factor $5{x}^{2}+10x$ , whose GCF is ${5x}$ :

## Factoring out binomial factors

The common factor in a polynomial does not have to be a monomial.

For example, consider the polynomial $x(2x-1)-4(2x-1)$ .

Notice that the binomial ${2x-1}$ is common to both terms. We can factor this out using the distributive property:

### Check your understanding

## Different kinds of factorizations

It may seem that we have used the term "factor" to describe several different processes:

- We factored monomials by writing them as a product of other monomials. For example,
.$12{x}^{2}=(4x)(3x)$ - We factored the GCF from polynomials using the distributive property. For example,
.$2{x}^{2}+12x=2x(x+6)$ - We factored out common binomial factors which resulted in an expression equal to the product of two binomials. For example:

While we may have used different techniques, in each case we are writing the polynomial as a

*of two or more factors. So in all three examples, we indeed***product***factored*the polynomial.## Challenge problems

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