Main content

## Algebra (all content)

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

Lesson 12: Factoring polynomials by taking common factors- Factoring with the distributive property
- 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)

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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

**GCF**(greatest common factor) of two or more monomials is the product of all their common prime factors. For example, the GCF of 6, x and 4, x, squared is 2, x.If 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, left parenthesis, b, plus, c, right parenthesis, equals, a, b, plus, a, c

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, squared and 4, x, plus, 3 as shown below:

Notice how each term in the binomial was multiplied by a common factor of start color #0c7f99, 3, x, squared, end color #0c7f99.

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

If we start with 3, x, squared, left parenthesis, 4, x, right parenthesis, plus, 3, x, squared, left parenthesis, 3, right parenthesis, we can use the distributive property to

*start color #0c7f99, 3, x, squared, end color #0c7f99 and obtain 3, x, squared, left parenthesis, 4, x, plus, 3, right parenthesis.***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, cubed, minus, 6, x, squared.

**Step 1: Find the GCF**

- 2, x, cubed, equals, start color #ca337c, 2, end color #ca337c, dot, start color #e07d10, x, end color #e07d10, dot, start color #e07d10, x, end color #e07d10, dot, x
- 6, x, squared, equals, start color #ca337c, 2, end color #ca337c, dot, 3, dot, start color #e07d10, x, end color #e07d10, dot, start color #e07d10, x, end color #e07d10

So the GCF of 2, x, cubed, minus, 6, x, squared is start color #ca337c, 2, end color #ca337c, dot, start color #e07d10, x, end color #e07d10, dot, start color #e07d10, x, end color #e07d10, equals, start color #0c7f99, 2, x, squared, end color #0c7f99.

**Step 2: Express each term as a product of start color #0c7f99, 2, x, squared, end color #0c7f99 and another factor.**

- 2, x, cubed, equals, left parenthesis, start color #0c7f99, 2, x, squared, end color #0c7f99, right parenthesis, left parenthesis, x, right parenthesis
- 6, x, squared, equals, left parenthesis, start color #0c7f99, 2, x, squared, end color #0c7f99, right parenthesis, left parenthesis, 3, right parenthesis

So the polynomial can be written as 2, x, cubed, minus, 6, x, squared, equals, left parenthesis, start color #0c7f99, 2, x, squared, end color #0c7f99, right parenthesis, left parenthesis, x, right parenthesis, minus, left parenthesis, start color #0c7f99, 2, x, squared, end color #0c7f99, right parenthesis, left parenthesis, 3, right parenthesis.

**Step 3: Factor out the GCF**

Now we can apply the distributive property to factor out start color #01a995, 2, x, squared, end color #01a995.

**Verifying our result**

We can check our factorization by multiplying 2, x, squared 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, squared, plus, 10, x, whose GCF is start color #0c7f99, 5, x, end color #0c7f99:

## Factoring out binomial factors

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

For example, consider the polynomial x, left parenthesis, 2, x, minus, 1, right parenthesis, minus, 4, left parenthesis, 2, x, minus, 1, right parenthesis.

Notice that the binomial start color #0c7f99, 2, x, minus, 1, end color #0c7f99 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, squared, equals, left parenthesis, 4, x, right parenthesis, left parenthesis, 3, x, right parenthesis.
- We factored the GCF from polynomials using the distributive property. For example, 2, x, squared, plus, 12, x, equals, 2, x, left parenthesis, x, plus, 6, right parenthesis.
- 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

## Want to join the conversation?

- i'm having trouble with this specific problem (our teacher didn't explain it well)

5m^2+21m-20=0

I am supposed to solve it using factoring, nothing else.(15 votes)- So first, we have a = 5, b = 21, and c = -20. ac = 5*-20 = - 100. Since this product is negative, I need two numbers which multiply to be -100 and add to be 21, but with a negative product, I can really say two numbers that multiply to be 100 and subtract to be 21, and since the 21 is positive, the bigger number has to be positive. So this is easy to guess: 25 - 4 = 21. Thus, we have the middle term broken into two parts, 21m = 25m - 4m. So our equation could be changed to 5m^2 +25m - 4m - 20 = 0. The GCF of the first two is 5m, and of the last two is -4, so 5m ( m + 5) - 4 (m + 5) = 0, we have a common factor of m + 5, so pulling that out, we get (5m - 4)(m + 5) = 0. Set the first equal to zero to get m = 4/5 and the second equal to zero to get m = -5.(53 votes)

- #7 even when I put in the answer given by the help button, it didn't work no matter how many times I tried it.(17 votes)
- It was fine when I did it. Use the controls underneath the answer box to assist you, they help.(10 votes)

- I seriously need help. I've learned factoring only once before today, and I still don't really get it. I mean I do, but I don't. I understand when the terms are broken down, like how 2x^2 is also 2*x*x. That's easy. It's mostly the finding the second term part, and making sure everything is in the right order. I know this is kind of a vague plea for assistance, but I don't know how else to explain my jumbled brain.(12 votes)
- Factoring is like breaking down two cakes and trying to take out the layers that are the same. lackluster analogy but the base of what you know (2x² = 2*x*x) is essential.

Let's start with a binomial, (4x²+8x)

The main part of factoring is to find the**G**reatest**C**ommon**F**actor. What common factors does 4x² and 8x? Breaking it down,**4**⋅**x**⋅x = 4x²

2⋅**4**⋅**x**= 8x

It seems that the biggest term that goes into both 4x² and 8x is 4 and x. Multiple 4 and x together to get 4x. It's also good to notice that the other x is 4x² is not included. That's because theres only 1 x in 8x. We would put the other x in if it were 8x², but it is not.

Let's now take out 4x from 4x²+8x :

4x(x+2) = 4x²+8x through the distributive property.

Trinomials are the same process but with more terms. hopefully that helps and if you want an example for trinomials lmk !(21 votes)

- So this isn't really on this page but I don't know where else to ask it. I have some homework that I don't fully understand a similar problem to the one I am confused on is 15x(x+6)^2+45x(x+6)+35(11 votes)
- On question 4, I typed in the wrong answer, then it said it was wrong. Then, I put in the right answer and it says it is still wrong! Why? Please fix!(10 votes)
- I have the following problem 6x^3 + 8x^2 - 4x but I can't get it right can someone help me, please.(8 votes)
- So let's begin with GCF. Looking at the polynomial, it seems that 2x is the GCF of that. Let's take that out:

2x(3x²+4x-2)

Noticing that there is a trinomial that**might**factor, we use the technique:

a * c = -6

a + c = 4

Noticing that all factors of 6 cannot add up to 4, we leave it at that. In some cases you cannot factor a trinomial, and this is an example of such.

Answer: 2x(3x²+4x-2)(7 votes)

- how do I factor x^2-36?(7 votes)
- x^2-36 factors into (x+6)(x-6). It's in the form a^2-b^2, which in factored form is (a+b)(a-b).(7 votes)

- The hint for number 7 is tricking me can you send a different hint(2 votes)
- A few problems are telling me to factor out multiple binomials in a single problem, such as x^2 - y^2 + 7x + 7y. I know the first step is to split the problem into multiple binomials (so that the problem then becomes (x^2 - y^2)+(7x + 7y), but I don't know how to make a GCF of two different variables with the same power. How would you be able to do that?(3 votes)
`x`

and`y`

don't have any common factors between themselves. But`x²-y²`

is a difference of squares:`x² - y² = (x - y)·(x + y)`

And`7x+7y`

have a common factor of 7:`7·x + 7·y = 7·(x + y)`

So now you have the following expression:`(x - y)·(x + y) + 7·(x + y)`

Notice that you can factor`(x + y)`

out:`(x + y)·((x - y) + 7) = (x + y)·(x - y + 7)`

And that's it.(8 votes)

- factors of (a+b)^3 - (a-b)^3(4 votes)
- Using the binomial theorem, we get

(𝑎 + 𝑏)³ − (𝑎 − 𝑏)³

= 𝑎³ + 3𝑎²𝑏 + 3𝑎𝑏² + 𝑏³ − (𝑎³ − 3𝑎²𝑏 + 3𝑎𝑏² − 𝑏³)

Combining like terms, we get

6𝑎²𝑏 + 2𝑏³

Finally, we factor out 2𝑏, which gives us

2𝑏(3𝑎² + 𝑏²)(4 votes)