If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

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.

Summary

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.
صرف 1 جواب چنو

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.
صرف 1 جواب چنو

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:
(ACa)x2+(BC+ADb)x+(BDc)
Now, let's define m=BC and n=AD.
(ACa)x2+(BCm+ADnb)x+(BDc)
According to this definition...
m+n=BC+AD=b
and
mn=(BC)(AD)=(AC)(BD)=ac
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!