﻿ 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;

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

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

CONCEPT QUESTIONS EXERCISE: