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Factoring quadratics: leading coefficient ≠ 1

Learn how to factor quadratic expressions as the product of two linear binomials. For example, 2x²+7x+3=(2x+1)(x+3).

What you need to know before taking this lesson

The grouping method can be used to factor polynomials with 4 terms by taking out common factors multiple times. If this is new to you, you'll want to check out our Intro to factoring by grouping article.
We also recommend that you review our article on factoring quadratics with a leading coefficient of 1 before proceeding.

What you will learn in this lesson

In this article, we will use grouping to factor quadratics with a leading coefficient other than 1, like 2x2+7x+3.

Example 1: Factoring 2x2+7x+3

Since the leading coefficient of (2x2+7x+3) is 2, we cannot use the sum-product method to factor the quadratic expression.
Instead, to factor 2x2+7x+3, we need to find two integers with a product of 23=6 (the leading coefficient times the constant term) and a sum of 7 (the x-coefficient).
Since 16=6 and 1+6=7, the two numbers are 1 and 6.
These two numbers tell us how to break up the x-term in the original expression. So we can express our polynomial as 2x2+7x+3=2x2+1x+6x+3.
We can now use grouping to factor the polynomial:
=  2x2+1x+6x+3=(2x2+1x)+(6x+3)Group terms=x(2x+1)+3(2x+1)Factor out GCFs=x(2x+1)+3(2x+1)Common factor!=(2x+1)(x+3)Factor out 2x+1
The factored form is (2x+1)(x+3).
We can check our work by showing that the factors multiply back to 2x2+7x+3.


In general, we can use the following steps to factor a quadratic of the form ax2+bx+c:
  1. Start by finding two numbers that multiply to ac and add to b.
  2. Use these numbers to split up the x-term.
  3. Use grouping to factor the quadratic expression.

Check your understanding

1) Factor 3x2+10x+8.
Choose 1 answer:

2) Factor 4x2+16x+15.

Example 2: Factoring 6x25x4

To factor 6x25x4, we need to find two integers with a product of 6(4)=24 and a sum of 5.
Since 3(8)=24 and 3+(8)=5, the numbers are 3 and 8.
We can now write the term 5x as the sum of 3x and 8x and use grouping to factor the polynomial:
= 6x2+3x8x4(1)=(6x2+3x)+(8x4)Group terms(2)=3x(2x+1)+(4)(2x+1)Factor out GCFs(3)=3x(2x+1)4(2x+1)Simplify(4)=3x(2x+1)4(2x+1)Common factor!(5)=(2x+1)(3x4)Factor out 2x+1
The factored form is (2x+1)(3x4).
We can check our work by showing that the factors multiply back to 6x25x4.
Take note: In step (1) above, notice that because the third term is negative, a "+" was inserted between the groupings to keep the expression equivalent to the original. Also, in step (2), we needed to factor out a negative GCF from the second grouping to reveal a common factor of 2x+1. Be careful with your signs!

Check your understanding

3) Factor 2x23x9.
Choose 1 answer:

4) Factor 3x22x5.

5) Factor 6x213x+6.

When is this method useful?

Well, clearly, the method is useful to factor quadratics of the form ax2+bx+c, even when a1.
However, it's not always possible to factor a quadratic expression of this form using our method.
For example, let's take the expression 2x2+2x+1. To factor it, we need to find two integers with a product of 21=2 and a sum of 2. Try as you might, you will not find two such integers.
Therefore, our method doesn't work for 2x2+2x+1, and for a bunch of other quadratic expressions.
It's useful to remember, however, that if this method doesn't work, it means the expression cannot be factored as (Ax+B)(Cx+D) where A, B, C, and D are integers.

Why is this method working?

Let's take a deep dive into why this method is at all successful. We will have to use a bunch of letters here, but please bear with us!
Suppose the general quadratic expression ax2+bx+c can be factored as (Ax+B)(Cx+D) with integers A, B, C, and D.
When we expand the parentheses, we obtain the quadratic expression (AC)x2+(BC+AD)x+BD.
Since this expression is equivalent to ax2+bx+c, the corresponding coefficients in the two expressions must be equal! This gives us the following relationship between all the unknown letters:
Now, let's define m=BC and n=AD.
According to this definition...
And so BC and AD are the two integers we are always looking for when we use this factorization method!
The next step in the method after finding m and n is to split the x-coefficient (b) according to m and n and factor using grouping.
Indeed, if we split the x-term (BC+AD)x into (BC)x+(AD)x, we will be able to use grouping to factor our expression back into (Ax+B)(Cx+D).
In conclusion, in this section we...
  • started with the general expanded expression ax2+bx+c and its general factorization (Ax+B)(Cx+D),
  • were able to find two numbers, m and n, such that mn=ac and m+n=b (we did so by defining m=BC and n=AD),
  • split the x-term bx into mx+nx, and were able to factor the expanded expression back into (Ax+B)(Cx+D).
This process shows why, if an expression can indeed be factored as (Ax+B)(Cx+D), our method will ensure that we find this factorization.
Thanks for pulling through!

Want to join the conversation?

  • leaf red style avatar for user Alemán
    So, this metod is not applicable to this equation: 3x^2+5x+10 for example. What method can I use to factorize this equation? Or there are any quadratics can´t be factorizated?
    (11 votes)
    Default Khan Academy avatar avatar for user
    • mr pink green style avatar for user David Severin
      The quadratic formula always works, so first look at the discriminant b^2 - 4 a c
      For your equation 5*2 - 4(3)(10) or 25 - 120 = - 95
      Since you cannot take SQRT of negative number in real domain
      This cannot be factorable since there is no real solution
      (63 votes)
  • mr pink red style avatar for user Vanessa Campos
    So on number 5, How do you know to group -4 with 6x^ or -9 with it?
    (6 votes)
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  • blobby green style avatar for user georgefabish
    On question 4)factor 3x^2 -2x- 5
    In working it out I got the answer (x-1)(3x+5) where as the correct answer was (3x-5)(x+1). The difference appears because there were multiple ways to split the GCF. As far as I can tell they both give the right answer and expand back into 3x^2-2x-5. Is this typical when solving these problems or did I do something wrong or is this just a random special case that happens to work?
    (5 votes)
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  • male robot donald style avatar for user aeppa
    I understand how to factor the quadratics but I don't really understand the part on how it works.
    (10 votes)
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  • aqualine seedling style avatar for user Mansi
    How would I solve a question like this: (2(1-x))/3 - 8/1=1/(6x)-(2x-3)/3
    (4 votes)
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  • male robot hal style avatar for user NIk NIk
    Why do we multiple 2 and 3 in the first example?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • duskpin tree style avatar for user Josie Cunningham
    What is the easiest way how to remember how to factor out quadratics?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • starky seedling style avatar for user 1000050035
      There are some songs on youtube to help memorize these formulas. I recommend the quadratic formula set to the tune of "Pop Goes the Weasel", because you can use that formula for a lot of different equations, and the song is very memorable. Hope this answers your question :)
      (2 votes)
  • mr pants orange style avatar for user Remedy
    A question about the first problem. My answer is "(x+2)(3x+4)", but there are no choices in there. Even if the values of a or b are opposite, is the answer the same?
    (1 vote)
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    • stelly blue style avatar for user Kim Seidel
      Your answer is correct (you can always check it by multiplying the 2 binomials). It matches option B in the answer list. Remember, the commutative property of multiplication tells us that 2(3) = 3(2). This property extends to any multiplication problem, including your binomials. (3x+4)(x+2) is the same as (x+2)(3x+4).

      Hope this helps.
      (7 votes)
  • male robot johnny style avatar for user hari the destroyer
    What if the resulting grouping ends up with contradicting signs inside the parenthesis (like 2x(x-4) -5(1+4) )

    This is just an example not the actual question I encountered.
    (3 votes)
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    • blobby green style avatar for user catterpylar
      I'm assuming you mean 2x(x-4)-5(x+4)
      You just change the sign of the number you are factoring out.
      For example, in the second part, you have -5(x+4). Multiplying it you get -5x-20. To get (x-4) you would factor out 5 instead of -5, to get 5(x-4)
      Now you have 2x(x-4)+5(x-4)
      which then becomes (2x+5)(x-4)
      You can also do this to the first part, 2x(x-4)
      switch 2x to -2x
      then you get -2x(x+4)-5(x+4) then (-2x-5)(x+4).
      (2 votes)
  • blobby green style avatar for user 117111
    How to add squares ?
    (3 votes)
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    • duskpin ultimate style avatar for user yaslinsy
      To answer this question I will use the example:
      4x^2 - 10x + 4
      When you take the constant from the squared variable in the front, and you multiply it by the last number, you get 16, and 16 is the result of 4^2, but you don't have to use 4 to get to sixteen. You could also use 8 and 2. -8 - 2 does equal -10, so it works and you get:
      4x^2 - 8x - 2x + 4
      then you group them
      (4x^2 - 8x) + (- 2x + 4)
      and factor it
      4x(x - 2) + -2(x - 2)
      factor it again to get your answer:
      (x - 2) (4x - 2)

      Hope this answered your question, tell me if it didn't
      (2 votes)