# 4.1 Factoring

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This module discusses how to solve quadratic equations by factoring.

When we multiply, we put things together: when we factor, we pull things apart. Factoring is a critical skill in simplifying functions and solving equations.

There are four basic types of factoring. In each case, I will start by showing a multiplication problem—then I will show how to use factoring to reverse the results of that multiplication.

## “pulling out” common factors

This type of factoring is based on the distributive property , which (as you know) tells us that:

$2x\left({4x}^{2}-7x+3\right)={8x}^{3}-\text{14}{x}^{2}+6x$

When we factor, we do that in reverse. So we would start with an expression such as ${8x}^{3}-\text{14}{x}^{2}+6x$ and say “Hey, every one of those terms is divisible by 2. Also, every one of those terms is divisible by $x$ . So we “factor out,” or “pull out,” a $2x$ .

${8x}^{3}-\text{14}{x}^{2}+6x=2x\left(\text{__}-\text{__}+\text{__}\right)$

For each term, we see what happens when we divide that term by $2x$ . For instance, if we divide ${8x}^{3}$ by $2x$ the answer is ${4x}^{2}$ . Doing this process for each term, we end up with:

${8x}^{3}-\text{14}{x}^{2}+6x=2x\left({4x}^{2}-7x+3\right)$

As you can see, this is just what we started with, but in reverse. However, for many types of problems, this factored form is easier to work with.

As another example, consider $6x+3$ . The common factor in this case is 3. When we factor a 3 out of the $6x$ , we are left with $2x$ . When we factor a 3 out of the 3, we are left with...what? Nothing? No, we are left with 1, since we are dividing by 3.

$6x+3=3\left(2x+1\right)$

There are two key points to take away about this kind of factoring.

1. This is the simplest kind of factoring. Whenever you are trying to factor a complicated expression, always begin by looking for common factors that you can pull out.
2. A common factor must be common to all the terms. For instance, ${8x}^{3}-\text{14}{x}^{2}+6x+7$ has no common factor, since the last term is not divisible by either 2 or $x$ .

## Factoring perfect squares

The second type of factoring is based on the “squaring” formulae that we started with:

${\left(x+a\right)}^{2}={x}^{2}+2\text{ax}+{a}^{2}$
${\left(x-a\right)}^{2}={x}^{2}-2\text{ax}+{a}^{2}$

For instance, if we see ${x}^{2}+6x+9$ , we may recognize the signature of the first formula: the middle term is three doubled , and the last term is three squared . So this is ${\left(x+3\right)}^{2}$ . Once you get used to looking for this pattern, it is easy to spot.

${x}^{2}+\text{10}x+\text{25}={\left(x+5\right)}^{2}$
${x}^{2}+2x+1={\left(x+1\right)}^{2}$

And so on. If the middle term is negative , then we have the second formula:

${x}^{2}-8x+\text{16}={\left(x-4\right)}^{2}$
${x}^{2}-\text{14}x+\text{49}={\left(x-7\right)}^{2}$

This type of factoring only works if you have exactly this case : the middle number is something doubled , and the last number is that same something squared . Furthermore, although the middle term can be either positive or negative (as we have seen), the last term cannot be negative.

All this may make it seem like such a special case that it is not even worth bothering about. But as you will see with “completing the square” later in this unit, this method is very general, because even if an expression does not look like a perfect square, you can usually make it look like one if you want to—and if you know how to spot the pattern.

## The difference between two squares

The third type of factoring is based on the third of our basic formulae:

#### Questions & Answers

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