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 [beautiful math coming... please be patient] $\,A^2 - B^2\,$ 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:

[beautiful math coming... please be patient] $\,(A + B)(A - B) = A^2 - AB + AB - B^2 = A^2 - B^2\,$
Thus, we have the following result:

FACTORING A DIFFERENCE OF SQUARES
For all real numbers [beautiful math coming... please be patient] $\,A\,$ and $\,B\,$: [beautiful math coming... please be patient] $$ A^2 - B^2 = (A + B)(A - B) $$
EXAMPLES:
Factor: [beautiful math coming... please be patient] $x^2 - 4$
Solution: [beautiful math coming... please be patient] $x^2 - 4 = x^2 - 2^2 = (x + 2)(x - 2)$
Factor: [beautiful math coming... please be patient] $9 - x^2$
Solution: [beautiful math coming... please be patient] $9 - x^2 = 3^2 - x^2 = (3 + x)(3 - x)$
Factor: [beautiful math coming... please be patient] $4x^2 - 9$
Solution: [beautiful math coming... please be patient] $4x^2 - 9 = (2x)^2 - 3^2 = (2x + 3)(2x - 3)$
Factor: [beautiful math coming... please be patient] $49x^2 - 64y^2$
Solution: [beautiful math coming... please be patient] $49x^2 - 64y^2 = (7x)^2 - (8y)^2 = (7x + 8y)(7x - 8y)$
Factor: [beautiful math coming... please be patient] $x^6 - 25$
Solution: [beautiful math coming... please be patient] $x^6 - 25 = (x^3)^2 - 5^2 = (x^3 + 5)(x^3 - 5)$
Factor: [beautiful math coming... please be patient] $x^2 - 5$
Solution:
Since [beautiful math coming... please be patient] $\,5\,$ 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:
[beautiful math coming... please be patient] $x^2 - 5 = x^2 - (\sqrt{5})^2 = (x + \sqrt{5})(x - \sqrt{5})$
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:   [beautiful math coming... please be patient] $\,\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$
Factor: [beautiful math coming... please be patient] $x^2 + 4$
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 [beautiful math coming... please be patient] $\,x^2 + 4\,$ can't even be factored using real numbers.
It can be factored if we're allowed to use numbers that aren't real:
[beautiful math coming... please be patient] $x^2 + 4 = x^2 - (2i)^2 = (x + 2i)(x - 2i)\,$, 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:
[beautiful math coming... please be patient] $x^6 + 64$ 
[beautiful math coming... please be patient] $\ \ = (x^3)^2 + 8^2$(so, it's a sum of squares)
[beautiful math coming... please be patient] $\ \ = (x^2)^3 + 4^3$(it's also a sum of cubes, which can be factored)
[beautiful math coming... please be patient] $\ \ = (x^2+4)(x^4 - 4x^2 + 16)$(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
Factor: