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### Course: Algebra (all content) > Unit 10

Lesson 15: Factoring quadratics by grouping# 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!

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