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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 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:
start color #0c7f99, 3, end color #0c7f99, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 4, x, plus, 3, right parenthesis, equals, start color #0c7f99, 3, end color #0c7f99, with, \overgroup, on top, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 4, x, right parenthesis, plus, start color #0c7f99, 3, end color #0c7f99, with, \overgroup, on top, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 3, right parenthesis
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!
start color #0c7f99, 3, end color #0c7f99, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 4, x, right parenthesis, plus, start color #0c7f99, 3, end color #0c7f99, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 3, right parenthesis, equals, start color #0c7f99, 3, end color #0c7f99, with, \overgroup, on top, with, \overgroup, on top, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 4, x, plus, 3, right parenthesis
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 factor out start color #0c7f99, 3, x, squared, end color #0c7f99 and obtain 3, x, squared, left parenthesis, 4, x, plus, 3, right parenthesis.
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

Problem 1
Write 2, x, left parenthesis, 3, x, right parenthesis, plus, 2, x, left parenthesis, 5, right parenthesis in factored form.
Choose 1 answer:
Choose 1 answer:

Factoring out the greatest common factor (GCF)

To factor the GCF out of a polynomial, we do the following:
  1. Find the GCF of all the terms in the polynomial.
  2. Express each term as a product of the GCF and another factor.
  3. 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.
start color #0c7f99, 2, end color #0c7f99, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, x, right parenthesis, minus, start color #0c7f99, 2, end color #0c7f99, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 3, right parenthesis, equals, start color #0c7f99, 2, end color #0c7f99, with, \overgroup, on top, with, \overgroup, on top, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, x, minus, 3, right parenthesis
Verifying our result
We can check our factorization by multiplying 2, x, squared back into the polynomial.
start color #0c7f99, 2, end color #0c7f99, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, x, minus, 3, right parenthesis, equals, start color #0c7f99, 2, end color #0c7f99, with, \overgroup, on top, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, x, right parenthesis, minus, start color #0c7f99, 2, end color #0c7f99, with, \overgroup, on top, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 3, right parenthesis
Since this is the same as the original polynomial, our factorization is correct!

Check your understanding

Problem 2
Factor out the greatest common factor in 12, x, squared, plus, 18, x.
Choose 1 answer:
Choose 1 answer:

Problem 3
Factor out the greatest common factor in the following polynomial.
10, x, squared, plus, 25, x, plus, 15, equals

Problem 4
Factor out the greatest common factor in the following polynomial.
x, start superscript, 4, end superscript, minus, 8, x, cubed, plus, x, squared, equals

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:
5, x, squared, plus, 10, x, equals, start color #0c7f99, 5, x, end color #0c7f99, left parenthesis, start fraction, 5, x, squared, divided by, start color #0c7f99, 5, x, end color #0c7f99, end fraction, plus, start fraction, 10, x, divided by, start color #0c7f99, 5, x, end color #0c7f99, end fraction, right parenthesis, equals, start color #0c7f99, 5, x, end color #0c7f99, left parenthesis, x, plus, 2, right parenthesis

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:
x, left parenthesis, start color #0c7f99, 2, x, end color #0c7f99, start color #0c7f99, minus, 1, end color #0c7f99, right parenthesis, minus, 4, left parenthesis, start color #0c7f99, 2, x, end color #0c7f99, start color #0c7f99, minus, 1, end color #0c7f99, right parenthesis, equals, left parenthesis, x, minus, 4, right parenthesis, left parenthesis, start color #0c7f99, 2, x, minus, end color #0c7f99, with, \overgroup, on top, with, \overgroup, on top, start color #0c7f99, 1, end color #0c7f99, right parenthesis

Check your understanding

Problem 5
Factor out the greatest common factor in the following polynomial.
2, x, left parenthesis, x, plus, 3, right parenthesis, plus, 5, left parenthesis, x, plus, 3, right parenthesis, equals

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:
x, left parenthesis, x, plus, 1, right parenthesis, plus, 2, left parenthesis, x, plus, 1, right parenthesis, equals, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, plus, 2, right parenthesis
While we may have used different techniques, in each case we are writing the polynomial as a product of two or more factors. So in all three examples, we indeed factored the polynomial.

Challenge problems

Problem 6
Factor out the greatest common factor in the following polynomial.
12, x, squared, y, start superscript, 5, end superscript, minus, 30, x, start superscript, 4, end superscript, y, squared, equals

Problem 7
A large rectangle with an area of 14, x, start superscript, 4, end superscript, plus, 6, x, squared square meters is divided into two smaller rectangles with areas 14, x, start superscript, 4, end superscript and 6, x, squared square meters.
The width of the rectangle (in meters) is equal to the greatest common factor of 14, x, start superscript, 4, end superscript and 6, x, squared.
What is the length and width of the large rectangle?
start text, W, i, d, t, h, end text, equals
meters
start text, L, e, n, g, t, h, end text, equals
meters

Want to join the conversation?

  • leafers seed style avatar for user Ed cole
    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.
    (14 votes)
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    • mr pink green style avatar for user David Severin
      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.
      (48 votes)
  • mr pink red style avatar for user jujumont333
    #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)
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  • starky ultimate style avatar for user Child
    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)
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    • leaf orange style avatar for user A/V
      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,
      4x ⋅x = 4x²
      2⋅4x = 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)
  • aqualine ultimate style avatar for user BethanyVemulapalli
    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)
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  • duskpin tree style avatar for user Grace Muriithi
    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)
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  • male robot hal style avatar for user Avery Baker
    I have the following problem 6x^3 + 8x^2 - 4x but I can't get it right can someone help me, please.
    (7 votes)
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    • leaf orange style avatar for user A/V
      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)
      (5 votes)
  • aqualine ultimate style avatar for user jacorijay
    how do I factor x^2-36?
    (6 votes)
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  • leafers seed style avatar for user Annie Giampietro
    The hint for number 7 is tricking me can you send a different hint
    (1 vote)
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  • primosaur tree style avatar for user Golden Pig
    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?
    (2 votes)
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    • piceratops tree style avatar for user VVCephei
      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.
      (7 votes)
  • blobby green style avatar for user Revika
    factors of (a+b)^3 - (a-b)^3
    (2 votes)
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