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## Algebra 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 2, x, squared, plus, 7, x, plus, 3.

## Example 1: Factoring $2x^2+7x+3$2, x, squared, plus, 7, x, plus, 3

Since the leading coefficient of left parenthesis, start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 7, end color #e07d10, x, start color #aa87ff, plus, 3, end color #aa87ff, right parenthesis is start color #11accd, 2, end color #11accd, we cannot use the sum-product method to factor the quadratic expression.
Instead, to factor start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 7, end color #e07d10, x, start color #aa87ff, plus, 3, end color #aa87ff, we need to find two integers with a product of start color #11accd, 2, end color #11accd, dot, start color #aa87ff, 3, end color #aa87ff, equals, 6 (the leading coefficient times the constant term) and a sum of start color #e07d10, 7, end color #e07d10 (the x-coefficient).
Since start color #01a995, 1, end color #01a995, dot, start color #01a995, 6, end color #01a995, equals, 6 and start color #01a995, 1, end color #01a995, plus, start color #01a995, 6, end color #01a995, equals, 7, the two numbers are start color #01a995, 1, end color #01a995 and start color #01a995, 6, end color #01a995.
These two numbers tell us how to break up the x-term in the original expression. So we can express our polynomial as 2, x, squared, plus, 7, x, plus, 3, equals, 2, x, squared, plus, start color #01a995, 1, end color #01a995, x, plus, start color #01a995, 6, end color #01a995, x, plus, 3.
We can now use grouping to factor the polynomial:
\begin{aligned}&\phantom{=}~~2x^2+1x+6x+3\\\\ &=({2x^2+1x}){+(6x+3)}&&\small{\gray{\text{Group terms}}}\\ \\ &=x({2x+1})+3({2x+1})&&\small{\gray{\text{Factor out GCFs}}}\\ \\ &=x(\maroonD{2x+1})+3(\maroonD{2x+1})&&\small{\gray{\text{Common factor!}}}\\\\ &=(\maroonD{2x+1})(x+3)&&\small{\gray{\text{Factor out } 2x+1}} \end{aligned}
The factored form is left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis.
We can check our work by showing that the factors multiply back to 2, x, squared, plus, 7, x, plus, 3.

### Summary

In general, we can use the following steps to factor a quadratic of the form start color #11accd, a, end color #11accd, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #aa87ff, c, end color #aa87ff:
1. Start by finding two numbers that multiply to start color #11accd, a, end color #11accd, start color #aa87ff, c, end color #aa87ff and add to start color #e07d10, b, end color #e07d10.
2. Use these numbers to split up the x-term.
3. Use grouping to factor the quadratic expression.

1) Factor 3, x, squared, plus, 10, x, plus, 8.

2) Factor 4, x, squared, plus, 16, x, plus, 15.

## Example 2: Factoring $6x^2-5x-4$6, x, squared, minus, 5, x, minus, 4

To factor start color #11accd, 6, end color #11accd, x, squared, start color #e07d10, minus, 5, end color #e07d10, x, start color #aa87ff, minus, 4, end color #aa87ff, we need to find two integers with a product of start color #11accd, 6, end color #11accd, dot, left parenthesis, start color #aa87ff, minus, 4, end color #aa87ff, right parenthesis, equals, minus, 24 and a sum of start color #e07d10, minus, 5, end color #e07d10.
Since start color #01a995, 3, end color #01a995, dot, left parenthesis, start color #01a995, minus, 8, end color #01a995, right parenthesis, equals, minus, 24 and start color #01a995, 3, end color #01a995, plus, left parenthesis, start color #01a995, minus, 8, end color #01a995, right parenthesis, equals, minus, 5, the numbers are start color #01a995, 3, end color #01a995 and start color #01a995, minus, 8, end color #01a995.
We can now write the term minus, 5, x as the sum of start color #01a995, 3, end color #01a995, x and start color #01a995, minus, 8, end color #01a995, x and use grouping to factor the polynomial:
\begin{aligned}&&&\phantom{=}~6x^2+\tealD{3}x\tealD{-8}x-4\\\\ \small{\blueD{(1)}}&&&=({6x^2+3x}){+(-8x-4)}&&\small{\gray{\text{Group terms}}}\\ \\ \small{\blueD{(2)}}&&&=3x({2x+1})+(-4)({2x+1})&&\small{\gray{\text{Factor out GCFs}}}\\ \\ \small{\blueD{(3)}}&&&=3x({2x+1})-4({2x+1})&&\small{\gray{\text{Simplify}}}\\ \\ \small{\blueD{(4)}}&&&=3x(\maroonD{2x+1})-4(\maroonD{2x+1})&&\small{\gray{\text{Common factor!}}}\\\\ \small{\blueD{(5)}}&&&=(\maroonD{2x+1})(3x-4)&&\small{\gray{\text{Factor out } 2x+1}}\\ \end{aligned}
The factored form is left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, 3, x, minus, 4, right parenthesis.
We can check our work by showing that the factors multiply back to 6, x, squared, minus, 5, x, minus, 4.
Take note: In step start color #11accd, left parenthesis, 1, right parenthesis, end color #11accd 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 start color #11accd, left parenthesis, 2, right parenthesis, end color #11accd, we needed to factor out a negative GCF from the second grouping to reveal a common factor of 2, x, plus, 1. Be careful with your signs!

3) Factor 2, x, squared, minus, 3, x, minus, 9.

4) Factor 3, x, squared, minus, 2, x, minus, 5.

5) Factor 6, x, squared, minus, 13, x, plus, 6.

## When is this method useful?

Well, clearly, the method is useful to factor quadratics of the form a, x, squared, plus, b, x, plus, c, even when a, does not equal, 1.
However, it's not always possible to factor a quadratic expression of this form using our method.
For example, let's take the expression start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 2, end color #e07d10, x, start color #aa87ff, plus, 1, end color #aa87ff. To factor it, we need to find two integers with a product of start color #11accd, 2, end color #11accd, dot, start color #aa87ff, 1, end color #aa87ff, equals, 2 and a sum of start color #e07d10, 2, end color #e07d10. Try as you might, you will not find two such integers.
Therefore, our method doesn't work for start color #11accd, 2, end color #11accd, x, squared, start color #e07d10, plus, 2, end color #e07d10, x, start color #aa87ff, plus, 1, end color #aa87ff, 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 left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis 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 a, x, squared, plus, b, x, plus, c can be factored as left parenthesis, start color #11accd, A, end color #11accd, x, plus, start color #e07d10, B, end color #e07d10, right parenthesis, left parenthesis, start color #1fab54, C, end color #1fab54, x, plus, start color #aa87ff, D, end color #aa87ff, right parenthesis with integers A, B, C, and D.
When we expand the parentheses, we obtain the quadratic expression left parenthesis, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, right parenthesis, x, squared, plus, left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, x, plus, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff.
Since this expression is equivalent to a, x, squared, plus, b, x, plus, c, the corresponding coefficients in the two expressions must be equal! This gives us the following relationship between all the unknown letters:
left parenthesis, start underbrace, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, end underbrace, start subscript, a, end subscript, right parenthesis, x, squared, plus, left parenthesis, start underbrace, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, end underbrace, start subscript, b, end subscript, right parenthesis, x, plus, left parenthesis, start underbrace, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff, end underbrace, start subscript, c, end subscript, right parenthesis
Now, let's define m, equals, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54 and n, equals, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff.
left parenthesis, start underbrace, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, end underbrace, start subscript, a, end subscript, right parenthesis, x, squared, plus, left parenthesis, start underbrace, start overbrace, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, end overbrace, start superscript, m, end superscript, plus, start overbrace, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, end overbrace, start superscript, n, end superscript, end underbrace, start subscript, b, end subscript, right parenthesis, x, plus, left parenthesis, start underbrace, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff, end underbrace, start subscript, c, end subscript, right parenthesis
According to this definition...
m, plus, n, equals, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, equals, b
and
m, dot, n, equals, left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, right parenthesis, left parenthesis, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, equals, left parenthesis, start color #11accd, A, end color #11accd, start color #1fab54, C, end color #1fab54, right parenthesis, left parenthesis, start color #e07d10, B, end color #e07d10, start color #aa87ff, D, end color #aa87ff, right parenthesis, equals, a, dot, c
And so start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54 and start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff 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 left parenthesis, b, right parenthesis according to m and n and factor using grouping.
Indeed, if we split the x-term left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, plus, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, x into left parenthesis, start color #e07d10, B, end color #e07d10, start color #1fab54, C, end color #1fab54, right parenthesis, x, plus, left parenthesis, start color #11accd, A, end color #11accd, start color #aa87ff, D, end color #aa87ff, right parenthesis, x, we will be able to use grouping to factor our expression back into left parenthesis, start color #11accd, A, end color #11accd, x, plus, start color #e07d10, B, end color #e07d10, right parenthesis, left parenthesis, start color #1fab54, C, end color #1fab54, x, plus, start color #aa87ff, D, end color #aa87ff, right parenthesis.
In conclusion, in this section we...
• started with the general expanded expression a, x, squared, plus, b, x, plus, c and its general factorization left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis,
• were able to find two numbers, m and n, such that m, n, equals, a, c and m, plus, n, equals, b left parenthesiswe did so by defining m, equals, B, C and n, equals, A, D, right parenthesis,
• split the x-term b, x into m, x, plus, n, x, and were able to factor the expanded expression back into left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis.
This process shows why, if an expression can indeed be factored as left parenthesis, A, x, plus, B, right parenthesis, left parenthesis, C, x, plus, D, right parenthesis, our method will ensure that we find this factorization.
Thanks for pulling through!

## Want to join the conversation?

• 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?
• 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
• So on number 5, How do you know to group -4 with 6x^ or -9 with it?
• It works either way.
• 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?
• Sorry, but your version creates the wrong sign on the middle term. You can see this if you multiply your factors. You get: 3x^2+5x-3x-5. Notice: 5x-3x = 2x, not -2x.

Hope this helps.
• Why do we multiple 2 and 3 in the first example?
• How would I solve a question like this: (2(1-x))/3 - 8/1=1/(6x)-(2x-3)/3
• Multiply both sides by 3, which gets -2x-22=(1/2x)-2x-3
Add 2x, so -22=(1/2x)-3. Then add 3, so -19=(1/2x). Multiply by 2x so -38x=1. Then divide by -38, so
x = (-1 / 38)
• I understand how to factor the quadratics but I don't really understand the part on how it works.
• What is the easiest way how to remember how to factor out quadratics?
• 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 :)
• How to add squares ?
• 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)
(x - 2) (4x - 2)

• With the example -3x^2+17x-20 I found that a=-3 b=20.

This was the working that followed:
(-3x^2-3x)+(20x-20)
=-3x(x+1)-20(x+1)
=(-3x-20)(x+1)

However, when I put this solution into the quiz it said I was wrong and in their solution, it substitutes a = 12 and b = 5.

Hence working would be:
(-3x^2+12x)+(5x-20)
=-3x(x+4)-5(x-4)
=(-3x+5)(x+4)

Does this have two solutions or have I done something wrong in my working out?
• Several issues, the first is that if you have (20x-20) and factor out a -20, you get -20(-x+1), and thus you do not have the same factor in x+1 and -x+1.
On the correct one, you start with -3x^2+ 12x which factoring out a -3x gives -3x(x-4). So to match, you should have factored out a 5 (not negative 5) to get 5(x-4) so that the x-4 matches to get (-3x+5)(x-4). If one has an x+4 and the other has x-4, you cannot combine them.
So if you expand by FOIL: (-3x+5)(x-4) = -3x^2 + 12x + 5x - 20 = -3x^2 + 17x - 20.
Expanding your first work gives (-3x-20)(x+1)= -3x^2 - 3x - 20x - 20 which gives 3x^2 - 23x - 20 (does not match). Expanding your second gives (-3x+5)(x+4) which gives -3x^2 - 12x + 5x + 20 = -3x^2 - 7x + 20 (does not match). So you struggled both with factoring out numbers in first case and factoring what cannot be factored in the second.
(1 vote)
• I can't get the idea of multiplying the constant and coefficient. An extra unatural method being added.Why not using the same method as the coefficient is 1?
2x^2 - 3x - 9
2x................-3
x.................3
(2x-3)(x+3)

suppose
ax^2 + bx + c
m.................p
n.................q

m * p + n * q = b
Step 1: come up with a pair of p,q.
Step 2: try out different pairs of m,n.
Step 3: done.
(1 vote)
• Technically, the method for when the coefficient is 1 is the same to when the coefficient is not 1. You just don’t acknowledge it because 1 times anything is always the same, so you just skip that part.