﻿ Factoring a Difference of Squares
FACTORING A DIFFERENCE OF SQUARES

Recall that factoring is the process of taking a sum/difference (things added/subtracted)
and renaming it as a product (things multiplied).

An expression of the form $\,A^2 - B^2\,$ is called a difference of squares.
It's a difference, because the last operation being performed is subtraction.
It's a difference of squares, because both $\,A^2\,$ and $\,B^2\,$ are squares.

Using FOIL:

$\,(A + B)(A - B) = A^2 - AB + AB - B^2 = A^2 - B^2\,$
Thus, we have the following result:

FACTORING A DIFFERENCE OF SQUARES
For all real numbers $\,A\,$ and $\,B\,$: $$A^2 - B^2 = (A + B)(A - B)$$
EXAMPLES:
Factor: $x^2 - 4$
Solution: $x^2 - 4 = x^2 - 2^2 = (x + 2)(x - 2)$
Factor: $9 - x^2$
Solution: $9 - x^2 = 3^2 - x^2 = (3 + x)(3 - x)$
Factor: $4x^2 - 9$
Solution: $4x^2 - 9 = (2x)^2 - 3^2 = (2x + 3)(2x - 3)$
Factor: $49x^2 - 64y^2$
Solution: $49x^2 - 64y^2 = (7x)^2 - (8y)^2 = (7x + 8y)(7x - 8y)$
Factor: $x^6 - 25$
Solution: $x^6 - 25 = (x^3)^2 - 5^2 = (x^3 + 5)(x^3 - 5)$
Factor: $x^2 - 5$
Solution:
Since $\,5\,$ is not a perfect square, this cannot be factored using integers.
Note that it can be factored if we're allowed to use non-integers:
$x^2 - 5 = x^2 - (\sqrt{5})^2 = (x + \sqrt{5})(x - \sqrt{5})$
In this exercise, you are factoring over the integers.
That is, you are to use only the integers for your factoring.
Recall that the integers are:   $\,\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$
Factor: $x^2 + 4$
Solution:
This can't be factored using integers.
Usually, a sum of squares can't be factored.

(☆ The remaining discussion is beyond the scope of Algebra I;
it is included for the benefit of more advanced readers.)

The expression $\,x^2 + 4\,$ can't even be factored using real numbers.
It can be factored if we're allowed to use numbers that aren't real:
$x^2 + 4 = x^2 - (2i)^2 = (x + 2i)(x - 2i)\,$, where $i^2 = -1$
In general, a sum of squares can't be factored.
However, a sum of squares might also be a sum of cubes, which is factorable, like this:
 $x^6 + 64$ $\ \ = (x^3)^2 + 8^2$ (so, it's a sum of squares) $\ \ = (x^2)^3 + 4^3$ (it's also a sum of cubes, which can be factored) $\ \ = (x^2+4)(x^4 - 4x^2 + 16)$ (use this: $\,A^3 + B^3 = (A + B)(A^2 - AB + B^2)\ \$)
So, you can't just make a blanket statement that sums of squares aren't factorable.
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Solving Simple Equations Involving Perfect Squares

CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.

In this exercise, you are factoring over the integers.
That is, you are to use only the integers for your factoring.
Thus, for this exercise, not factorable means not factorable over the integers.
PROBLEM TYPES:
 1 2 3 4 5 6 7 8 9 10 11 12
AVAILABLE MASTERED IN PROGRESS
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