Explore Quadratic equations

# Factoring

Just because math is getting more complicated doesn’t mean we can’t make it a little easier for ourselves! Factoring is one way we can make our calculations a little clearer and easier to follow.

And who wouldn’t want to make math easier?

## What does it mean to factor an expression?

Factoring an expression means rewriting it as a product of two or more factors.

Really, it’s another shining moment for the distributive property of multiplication! We’re not using it to calculate the product, but we can use it to write an expression as a product.

Factoring is often used when solving quadratic equations because, then, the quadratic expression is rewritten as a product of linear expressions. Then, by remembering the zero product property, we know the solutions of the equation are the solutions of linear equations.

“Wait…what is a quadratic equation?!’’

Good question! A quadratic equation is an equation in which the variable is raised to the second power. (Click that link if you need more context!)

As a reminder, a quadratic equation written in standard form looks like this:

$$ax^2+bx+c=0$$

### Why is factoring so useful?

Factoring is used for solving quadratic equations. Some expressions can easily be factored so that the quadratic expression is rewritten as the product of two linear expressions, which we all know how to solve!

The most common factoring is using the square of the sum, the square of the difference, the difference of the squares, and factoring out the common factor.

## How to factor

Now that we know what the factoring is and why it’s useful, it’s time to see it in action! Let’s go through some problems together.

### Example 1

Solve a quadratic equation using factoring:

$$25x^{2} - {1} = 0$$

We know that $$1$$ can be rewritten as $$1^2$$, so let’s rewrite the number:

$${25x^{2}} – {1^2} = 0$$

Remember the difference of squares formula $${a^2}-{b^2}=({a}-{b})({a}+{b})$$? We’re about to use it on our expression:

$$({5x}-{1})({5x}+{1})=0$$

When the product of factors equals $$0$$, at least one factor is $$0$$. So, separate the equation by cases:

$$5x{-}{1}=0$$

$$5x{+}{1}=0$$

Move the constants to the right-hand side and change their signs:

$$5x=1$$

$$5x={-}1$$

Divide both sides of the equations by $$5$$:

$$x=\frac15$$

$$x=-\frac15$$

Nice! Our equation has two solutions:

$$x_1=\frac15, ~x_2=-\frac15$$

That wasn’t awful, right? Let’s do another one.

### Example 2

Solve a quadratic equation using factoring:

$${t^2}-{4t}+{4}=0$$

Notice that $$4=2^2$$ and that $$4t$$ can be rewritten as $$2\times2\times t$$:

$${t^2}-2\times{2}\times {t}+{2^2}=0$$

Let’s use the square of the difference formula $${a^2}-2{a}{b}+{b^2}=({a}-{b})^2$$ on our expression:

$$({t}-{2})^2=0$$

The only way a power can be $$0$$ is when the base is equal to $$0$$, so:

$$t{-}{2}=0$$

Move the constant to the right-hand side and change its sign:

$$t=2$$

This equation only has one solution:

$$t=2$$

We did it again!

Let’s review the process so you can learn to use it with any problem:

## Study summary

1. Factor the expression.
2. Separate into possible cases.
3. Solve the equations to get the solutions!

### Do it yourself!

Having fun factoring? Just want some extra practice before working on something graded? Lucky for you, we’ve compiled some practice problems to help you hone your skills!

Solve quadratic equations using factoring:

1. $$x^2-8x+16=0$$
2. $$9t^2+6t+1=0$$
3. $$4x^2=49$$
4. $$16x^2-4=0$$

Solutions:

1. $$x=4$$
2. $$t=-\frac{1}{3}$$
3. $$x_1=-\frac{7}{2}, ~x_2=\frac{7}{2}$$
4. $$x_1=-\frac{1}{2}, ~x_2=\frac{1}{2}$$

If you’re struggling through the solving process, take a breath! That’s totally normal and okay. When you’re not sure how to proceed, you can always scan the problem using your Photomath app so that we can walk you through to the other side!

Here’s a sneak peek of what you’ll see:

/

## Got algebra homework?

Head over to the Photomath app for instant, step-by-step solutions to all of your algebra problems.