## Difference of Squares Lesson

### The Difference of Squares Pattern

There are special patterns that some polynomials follow. **The difference of squares pattern is when a polynomial is made up of a square minus another square.** This pattern allows us to easily factor the polynomial. The formula is given as:*a ^{2} - b^{2} = (a + b)(a - b)*

Where *a* and *b* may be any algebraic expression. For example, *a* could be a variable such as "y", and *b* could be a number such as "5". Or, *a* could be a more complicated expression such as "2x^{3}". The difference of squares pattern can be used for any set of algebraic expressions.

#### When to use the Difference of Squares

The difference of squares pattern comes in handy when we must factor a polynomial to find the roots/zeros or to simplify it. For example, we might be tasked with finding the roots/zeros of the equation y = x^{2} - 9.

We can use the difference of squares to factor the right-hand side of the equation and then find the roots/zeroes. Applying the difference of squares pattern to x^{2} - 9 gives us (x + 3)(x - 3). Using this factored form, we can determine that the roots/zeros are x = -3 and x = 3.

### Difference of Squares Example Problems

Let's go through a couple of example problems together to practice using the difference of squares pattern.

#### Example Problem 1

Using the difference of squares pattern, factor the polynomial x^{2} – 49.

Solution:

- Let's set a = x and b = 7 because a
^{2}= x^{2}and b^{2}= 49. - Applying the pattern (a + b)(a - b), we get (x + 7)(x - 7).
**The factored form of the polynomial x**^{2}– 49 is (x + 7)(x - 7).

#### Example Problem 2

Find the x-intercepts of y = 4x^{4} - 64.

Solution:

- We find the x-intercepts by setting y = 0. This gives us 0 = 4x
^{4}- 64. - We must now find the zeroes of the expression 4x
^{4}- 64. The expression is a difference of squares binomial. - Applying the formula, we get:

a^{2}- b^{2}= (a + b)(a - b)

4x^{4}- 64 = (2x^{2}+ 8)(2x^{2}- 8) - Now that it's in factored form, we can find the zeroes. (2x
^{2}+ 8)(2x^{2}- 8) has zeroes at x = 2 and x = -2. **The x-intercepts are located at x = 2 and x = -2.**