If a

polynomial *f(x)* is divided by

*x - a*, the

remainder will be

*f(a)*. A polynomial has the

factor *x - a* if, and only if

*f(a) = 0*.

From these statements we can derive a rather interesting way to factor polynomials, namely polynomial division, which is based on long division but using polynomials.

That doesn't sound very nice, does it? It isn't that bad actually, except for having to relearn the long since forgotten skill of long division. The principle is just the same, to "eliminate" term by term until you reach a stop, at which point you may or may not have a remainder. An example might serve to illustrate this before I elaborate further. Note that you may use whichever method of long division you're most comfortable with.

2x^3 - 3x^2 - 27
----------------
x - 3
2x^2 + 3x + 9
-------------------------------
2x^3 - 3x^2 + 0x - 27 | x - 3
- 2x^3 - 6x^2 --------
----------------------
3x^2 + 0x - 27
3x^2 - 9x
---------------
- 9x - 27
- 9x - 27
----------
0 <---- remainder
Verify the remainder: f(3) = 2 * 3^3 - 3 * 3^2 - 27 = 54 - 27 - 27 = 0

As you can see you have to think "by what do I need to multiply (x - 3) by to eliminate (2x^3)" first, to which the answer is (2x^2). Write that down on top, do the multiplication, and subtract the result from the function. The next step is to get rid of (3x^2), so you'll need to multiply (x - 3) by (3x). Repeat until you have a remainder that you cannot eliminate.

This division shows the following:

2x^3 - 3x^2 - 27 = (2x^2 + 3x + 9)(x - 3)

Now that second equation certainly looks easier to solve (unless you simply typed the first one into your calculator, you lazy bum!) This can help solve otherwise hard or impossible polynomial equations if you, one way or another, know one of the solutions. For example, if you have been given the following data:

*f(x) = 0*
*f(x) = x^3 + 8x - 24*
*f(2) = 0*

Simply factor

*f(x)* by

*x - 2* using polynomial division, and you'll get

*(x - 2)(x^2 + 2x + 12)*, which can be split into two equations, together

yielding all three solutions.