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2 Tutorials that teach Quadratic Factoring using the Difference of Squares

# Quadratic Factoring using the Difference of Squares

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Author: Colleen Atakpu
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In this lesson, students will learn how to factor quadratic expressions by using the difference of two squares.

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Tutorial

## Video Transcription

Let's go over our objectives for today. We'll start by looking at quadratic expressions. We'll then look at a specific type of quadratic expression called the difference of squares. We'll look at how to factor using the difference of squares. And finally, we'll do some examples factoring difference of square expressions.

Let's start by looking at quadratic expressions. A quadratic expression is a polynomial that can be written in the form ax squared plus bx plus c. In this expression, the first term, ax squared, is called the x squared term, and the coefficient is a. The second term, bx, is called the x term, and the coefficient is b. And the third term, c, is called the constant term.

Now let's look at a specific type of quadratic expression called the difference of squares. Here's an example of a quadratic expression, x squared minus 16. We know that this is quadratic because the highest exponent in the expression is 2. We also notice that the constant term, 16, is a perfect square, because the square root of 16 is 4, which is an integer.

So we can rewrite our expression as x squared minus 4 squared. And this is what we call a difference of square expression, because there are two values squared, x and the integer, with subtraction between the two. Difference of squares expressions can be factored similarly to other quadratic expressions. So let's see what that looks like.

Now let's look at how to factor difference of squares expressions using our previous example. We want to factor the expression x squared minus 16. To factor, we need to find two numbers that multiply to the constant term, and add to the coefficient of the middle term.

When the x term is absent from a quadratic expression, the coefficient of the x term is 0. So we can write the x term with the coefficient of 0, but because anything multiplied by 0 is 0, the entire term is 0. So we don't need to write anything.

Now when we factor, the only pair of numbers that sum to 0 are opposites of each other. For example, positive 1 and negative 1 sum to 0 and are opposites of each other. And negative 5 and positive 5 also sum to 0 and are opposites. Again, notice that the only difference between the numbers is their sign.

Going back to our expression that we want to factor, we also notice that our constant term, 16, is a perfect square. So we can now write our expression as x squared minus 4 squared. To factor our expression, we need to find two numbers that add to 0 and multiply to negative 16. 4 and negative 4 are opposites, so they sum to 0, and they also multiply to negative 16. Therefore, we can factor our expression as x plus 4 times x minus 4.

Notice that the integer being squared in the original expression, 4, is part of both factors. In the first factor, x plus 4, and in the second factor, x minus 4. Therefore, in general, we can factor difference of square equations as a squared minus b squared is equal to a plus b times a minus b.

Here's your last example. We want to factor the expression 9x squared minus 25. We notice that 9 and 25 are perfect squares. We can use the difference of squares formula to factor the expression. The square root of 9 is 3, and the square root of 25 is 5, so we can rewrite our expression as 3x squared minus 5 squared.

Now we can use our formula to factor the expression. Our a term in the formula is 3x and our b term in the formula is 5. So to factor, we have 3x plus 5 times 3x minus 5.

We can verify that we have factored the expression correctly by performing FOIL to see if we get our original expression. So to multiply the two expressions together, we multiply our first terms, 3x and 3x, which is 9x squared. Our outside terms, 3x times negative 5, is negative 15x. Our inside terms, 5 and 3x, multiplied, give us positive 15x. And our last terms, 5 and negative 5, multiplied give us negative 25.

We can combine our middle terms, which are like terms, which cancel each other out, to give us 9x squared minus 25. So we did factor our expression correctly.

Let's go over our important points from today. Make sure you get them in your notes, so you can refer to them later.

In a quadratic expression, the highest exponent in the expression is 2. In difference of squares expressions, there are two values squared, x and the integer, with subtraction between them. The goal of factoring is to find two numbers that multiply to the constant term, and add to the coefficient of the middle term. And when factoring difference of squares expressions, the coefficient of the middle term is 0, and the only pair of numbers that sum to 0 are opposites of each other.

So I hope these important points and examples helped you understand a little bit more about quadratic factoring using the difference of squares. Keep using your notes, and keep on practicing, and soon you'll be a pro. Thanks for watching.

## Notes on "Quadratic Factoring using the Difference of Squares"

00:00 – 00:37 Introduction