*I wanted to do this before the Algebra Regents exams when it might have been more helpful to some students, but the fact is, they don't actually follow me. They just find the posts with the questions and answers on them through search engines.*

The **Difference of Squares** is a common topic for **Algebra 1**, but what about other powers, such as cubes and quarts? (Quarts are is "quartic", fourth power.)

## Squares and Quarts

In a binomial that looks like ** x^{2} - n^{2}**, where

*n*is any number, which will be squared, we can factor the two terms into conjugates:

*(x + n)(x - n)*All nice and neat. When you multiply the two conjugates *(which have the same terms, but one has a plus and one has a minus)*, the two *x* terms -- (* nx*) and (

*) -- are additive inverses and will sum to zero.*

**-nx**Every now and then, state exams like to throw a curve ball at you, and use a higher power, usually 4.

What if the problem said:

*Factor completely*

**x**^{4}- 16Since 4 is an even power, x

^{4}is a perfect square. And, of course, 16 is a perfect square. So the regular rule applies:

*=*

**x**^{4}- 16

*(x*^{2}+ 4)(x^{2}- 4)Ah, but have we factored "completely" as the question asked? We can't do anything with the first factor, (x

^{2}+ 4), which has no real roots, but what about the other factor? It's another

*Difference of Squares*, so we can apply the rule again:

*=*

**x**^{4}- 16**=**

*(x*^{2}+ 4)(x^{2}- 4)

*(x*^{2}+ 4)(x + 2)(x - 2)So, if they really wanted to be mean, er, I mean "challenging", they could go another step further and ask:

*Factor completely*

**x**^{8}- 256## Sum of Cubes

I know that I started by saying "difference of ...", but that was to keep everything flowing.

Unlike squares (and other even powers), you can factor the *sum* of two cubes. Also unlike squares, a perfect cube can be a negative number, so it also could be written as a difference of cubes. So I'm not wrong -- I just renamed it for no reason.

There's a simple format for the * Sum of Two Cubes*:

**=**

*(x*^{3}+ n^{3})

*(x + n)(x*^{2}- nx + n^{2})First thing you should see is that (x + n) means that there is a root at ** x = -n**.

The second thing you should see is that that is the only real root. If you check the discriminant, b^{2} - 4ac, you get the following:

(n)^{2} - 4(1)(n^{2}).

As long as *n* has a nonzero value, the discriminant will always be less than zero. This example had a leading coefficient of 1, but it is still true if there is a different number in front the x^{3} term.

### Final Example

If we wanted to put this together, we could try to factor ** x^{6} - 64**.

Or really put it all together and got with an ** x^{12}** term. But who wants to do that?

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