`a`_{n}x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2}. . . a_{1}x + a_{0}

The degree of a polynomial is the highest total of powers of variables (x, y, etc.) of a single term, so in the polynomial `2xy`^{2} + x^{2} the degree is three (in the first term, x has a power of one). The standard form of a polynomial is when you write it with the degrees descending (`x`^{2} + x + 3, not `x + x`^{2} + 3)

To factor a polynomial (If you already know how to then skip down to the AC method. You'll like it. A lot.) you first factor out the common factor, if there is one, using the distributive property:

**Ex 1)** `2x`^{2} + 4x = 2x(x + 2)

**Ex 2)** `2x`^{2} + 6x + 8 = 2(x^{2} + 3x + 4)

With a binomial (two terms, as in **Ex 1**) that's all. If you have a trinomial (three terms, as in **Ex 2**) you're just getting started.

You usually have to find two binomials (B_{1} and B_{2}) whose first terms multiply to the first term of your trinomial, last terms multiply to the last term of the trinomial, and B_{1}'s first term times B_{2}'s last term plus vice versa equals the middle term (FOIL users: Inside + Outside=Middle)

**Ex 3)** `x`^{2} + 3x + 2 = (x + 1)(x + 2)

If the first term of your trinomial has a coefficient (a) of 1--as shown above--then the first terms of the binomials are x. Otherwise, you have to play around searching for the proper factors to get it right. That's where the following method comes in:

**The AC Method**

First factor out the common factor. Always, always, always do this.

Now you have `ax`^{2} + bx + c a isn't 1.

Change it to `x`^{2} + bx + ac. (If you're stuck wondering how the hell to move the *a* all the way over to the *c*, don't bother. Just do it.)

Factor `x`^{2} + bx + ac into your (presumably) two binomials. Then stick *a* back into the first terms of both of them, factor out the common factor and toss it out. You're done.

**Ex 4)** `6x`^{2} + 2x-4

`2(3x`^{2} + x-2) (Factor out common factor)

`2(x`^{2} + x-6) (move *a* to third term)

`2(x + 3)(x-2)` (factor)

`2(3x + 3)(3x-2)` (put *a* back into first terms)

`2(x + 1)(3x-2)` (factor out and delete common factor)

If you're planning on using the AC Method a lot you may want to work on your factoring large numbers because *ac* is often rather large.

Now, I know you're thinking, "What if I have a four-term (or more) polynomial?" Easy: Take a few terms, and slap parenthesis around them (Hint, put together terms that have common factors or that look like they'll factor easily.)

**Ex 5)** `2x`^{3} - 3x^{2} + 4x - 6

`(2x`^{3} - 3x^{2}) + (4x - 6)

`x`^{2}(2x - 3) + 2(2x - 3)

`(x`^{2} + 2)(2x - 3)

That last example (first and last steps anyway) was taken from *College Algebra* by Michael Sullivan because I was having a heck of a time making up a good example. (I'm always coming up with prime polynomials in my example and having to modify them so I can factor them. I wish my math teacher had let me do that in my homework.)

Now you need to do some heavy memorising. These are special polynomials and how to factor them. Knowing how to recognise them will help you enormously, both in multiplication and factoring:

Difference of Squares: `x`^{2} - a^{2} = (x - a)(x + a) (**Ex 6)** `x`^{2} - 144 = (x + 12)(x - 12))

Perfect Squares: `x`^{2} ± 2ax + a^{2} = (x ± a)^{2}

Unnamed, but bears remembering: `x`^{2} (a + b)x + ab = (x + a)(x + b)

Unnamed, but bears remembering: `acx`^{2} + (ad + bc)x + bd = (ax + b)(cx + d)

Perfect Cubes: `x`^{3} + 3ax^{2} + 3a^{2}x + a^{3} = (a + x)^{3}, `x`^{3} - 3ax^{2} + 3a^{2}x - a^{3} = (a - x)^{3}

Sum of Two Cubes: `x`^{3} + a^{3} = (x + a)(x^{2} - ax + a^{2})

Difference of Two Cubes: `x`^{3} - a^{3} = (x - a)(x^{2} + ax +a^{2})

Take the coefficients of (x + y)^{n} and look at the *n*th row of Pascal's Triangle (the "1" at the top is 0th). Cute *and* useful. (m_turner is smart.)

Please /msg with additions and/or corrections.

This writeup is licensed under the GNU Free Documentation License.