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## College Algebra

### Course: College Algebra > Unit 4

Lesson 4: Factoring quadratics by grouping- Intro to grouping
- Factoring by grouping
- Factoring quadratics by grouping
- Factoring quadratics: leading coefficient ≠ 1
- Factor quadratics by grouping
- Factoring quadratics: common factor + grouping
- Factoring quadratics: negative common factor + grouping

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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 $2{x}^{2}+7x+3$ .

## Example 1: Factoring $2{x}^{2}+7x+3$

Since the leading coefficient of $({2}{x}^{2}{+7}x{+3})$ is ${2}$ , we cannot use the sum-product method to factor the quadratic expression.

Instead, to factor ${2}{x}^{2}{+7}x{+3}$ , we need to find two integers with a product of ${2}\cdot {3}=6$ (the leading coefficient times the constant term) and a sum of ${7}$ (the $x$ -coefficient).

Since ${1}\cdot {6}=6$ 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
$2{x}^{2}+7x+3=2{x}^{2}+{1}x+{6}x+3$ .

We can now use grouping to factor the polynomial:

The factored form is $(2x+1)(x+3)$ .

We can check our work by showing that the factors multiply back to $2{x}^{2}+7x+3$ .

### Summary

In general, we can use the following steps to factor a quadratic of the form ${a}{x}^{2}+{b}x+{c}$ :

- Start by finding two numbers that multiply to
and add to${a}{c}$ .${b}$ - Use these numbers to split up the
-term.$x$ - Use grouping to factor the quadratic expression.

### Check your understanding

## Example 2: Factoring $6{x}^{2}-5x-4$

To factor ${6}{x}^{2}{-5}x{-4}$ , we need to find two integers with a product of ${6}\cdot ({-4})=-24$ and a sum of ${-5}$ .

Since ${3}\cdot ({-8})=-24$ and ${3}+({-8})=-5$ , the numbers are ${3}$ and ${-8}$ .

We can now write the term $-5x$ as the sum of ${3}x$ and ${-8}x$ and use grouping to factor the polynomial:

The factored form is $(2x+1)(3x-4)$ .

We can check our work by showing that the factors multiply back to $6{x}^{2}-5x-4$ .

**Take note:**In step

### Check your understanding

## When is this method useful?

Well, clearly, the method is useful to factor quadratics of the form $a{x}^{2}+bx+c$ , even when $a\ne 1$ .

However, it's not always possible to factor a quadratic expression of this form using our method.

For example, let's take the expression ${2}{x}^{2}{+2}x{+1}$ . To factor it, we need to find two integers with a product of ${2}\cdot {1}=2$ and a sum of ${2}$ . Try as you might, you will not find two such integers.

Therefore, our method doesn't work for ${2}{x}^{2}{+2}x{+1}$ , 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 $(Ax+B)(Cx+D)$ where $A$ , $B$ , $C$ , and $D$ are integers.

**cannot**be factored as## 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 $a{x}^{2}+bx+c$ can be factored as $({A}x+{B})({C}x+{D})$ with integers $A$ , $B$ , $C$ , and $D$ .

When we expand the parentheses, we obtain the quadratic expression $({A}{C}){x}^{2}+({B}{C}+{A}{D})x+{B}{D}$ .

Since this expression is equivalent to $a{x}^{2}+bx+c$ , 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 $m={B}{C}$ and $n={A}{D}$ .

According to this definition...

and

And so ${B}{C}$ and ${A}{D}$ 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 $({B}{C}+{A}{D})x$ into $({B}{C})x+({A}{D})x$ , we will be able to use grouping to factor our expression back into $({A}x+{B})({C}x+{D})$ .

In conclusion, in this section we...

- started with the general expanded expression
and its general factorization$a{x}^{2}+bx+c$ ,$(Ax+B)(Cx+D)$ - were able to find two numbers,
and$m$ , such that$n$ and$mn=ac$ $m+n=b$ we did so by defining$($ and$m=BC$ ,$n=AD)$ - split the
-term$x$ into$bx$ , and were able to factor the expanded expression back into$mx+nx$ .$(Ax+B)(Cx+D)$

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?

- 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)
- 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)

- So on number 5, How do you know to group -4 with 6x^ or -9 with it?(6 votes)
- It works either way.(3 votes)

- 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)- 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.(15 votes)

- I understand how to factor the quadratics but I don't really understand the part on how it works.(10 votes)
- How would I solve a question like this: (2(1-x))/3 - 8/1=1/(6x)-(2x-3)/3(4 votes)
- 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)(3 votes)

- Why do we multiple 2 and 3 in the first example?(7 votes)
- What is the easiest way how to remember how to factor out quadratics?(4 votes)
- 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)

- 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)
- 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)

- 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)- 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)

- How to add squares ?(3 votes)
- 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)