How to factor these equations

1. Feb 12, 2007

I dont know why but I have a lot of trouble factoring. After 3 years of it I would expect to be a lot better at it but I still find it as hard as when I first learned it. Im just wondering if anyone here can help me understand it a little different then my teachers have.

To date I think I finally understand stuff like
$$2x^2-2x-12$$
$$x^3-5x^2-6x$$
$$5x^3-45x$$

The stuff I am having trouble with is things like this
$$x^3+x^2-4x-4$$
$$x^8-1$$
$$2m(m-n) + (m+n)(m-n)$$

Is there any certain rules or guidelines I may be able to follow to make these kind of questions easier. Right now im not really sure how to start them.

2. Feb 12, 2007

cristo

Staff Emeritus
For the first one if you can't factor it, then guess a factor and divide the polynomial by that factor to get a quadratic.

The second one seems pretty straightfoward... what have you tried?

In the third one, what is the common factor in both terms?

3. Feb 12, 2007

The first one I understand now, I get (x+1)(x+2)(x-2)

The second one im still not really sure

Here is the third one I think
(m-n)[2m + (m+n)]
(3m+n)(m-n)

4. Feb 12, 2007

rock4christ

I HATE factoring. I am great at all math I have ever tried(literally) and got a D on a factoring test. any thing beyond the complexity of x2 +ax +b i fail at. once x2 has a coefficient I fail miserably

5. Feb 12, 2007

turdferguson

In order to understand the 3rd line better, Ill explain another way to factor the 1st line. You can group x^3 with x^2 and -4x with -4. Then you can factor each piece and youre left with:
x^2(x+1) -4(x+1)
This form is very similar to that third line. Next, you treat x^2 and -4 as coeffieients and factor out an (x+1) from both terms:
(x+1)(x^2 - 4) This can then be reduced to what you got, and the third line also looks good

But youre having the most trouble with the second line, x^8 - 1. Both terms are perfect squares. Whats the square root of x^8? Are you done after that, or can it be factored further?

6. Feb 13, 2007

AlephZero

For problems where you know there is an easy answer (otherwise you wouldn't have been asked the question!) this often helps:

If you want to factor $ax^n \pm \dots \pm b$ then for a factor of the form $px \pm q$ it is a good bet that p is a factor of a, and q is a factor of b.

Also, use the remainder theorem: if (x-a) is a factor, the polynomial is zero when x = a. So in the $x^3+x^2-4x-4$ example it's fairly obvious the polynomial is zero when x = -1 therefore (x+1) is a factor.