How do you show that a polynomial has a double factor?

Question

Would anyone be able to help me with this question?

For the polynomial:

P(x)=x^5-x^4-10x^3+2x^2+16x-8, I have been told that two factors are (x-1) and (x+2) and I have to show that one of these is a double factor...how do you do that? I know that a double factor means that it's a perfect square (i.e. (x-1)^2 or (x+2)^2) but that's all...I would really appreciate some help...can anyone explain this to me? Or show me a website that will? Thanks so much! :D

Answer 1

You have to know how to divide polynomials. Actually, for this sort of thing, there's an easy way called synthetic division, or Ruffini's Rule.

So what you do is divide those two factors you were given, then try each one again and see which one works.

Basically it's just a lot of adding and carrying over. It takes a bit of practice to get good at this, but that's how it's done. The number you start with on the left is actually the opposite of what you might expect. If you want to test (x - 1) as a factor, start with a positive 1 on the left. To check (x + 2), start with -2.

http://en.wikipedia.org/wiki/Synthetic_division

Answer 2

Synthetic division.

You need to divide x^5-x^4-10x^3+2x^2+16x-8 by Divide by x^2-2x+1 and x^2 +4x + 4. If they go in evenly, you'll know it's a double factor.

First let's check (x - 1)^2 -> x^2 - 2x + 1
It get x^3 + x^2 - 9x + 17 with a remainder of 25.. this isn't it.

Now let's check (x+2)^2 = x^2 + 4x + 4
You get x^3 - 5x^2 + 6x -2 without any remainder. x+2 wins!

Answer 3

A polynomial p(x) has a double root at x=a if and only if both p(a) and p'(a)=0.

In your case, P'(x)=5x^4-4x^3-30x^2+4x+16. Are either 1 or -2 a root?