# Factoring a polynomial

• Mar 27th 2013, 08:52 PM
ReneG
Factoring a polynomial
I need help factoring $x^5+2x^3-x^2-2$

I tried it by grouping $(x^5 - x^2) + (2x^3 -2)$

Then I got $x^2(x^3 - 1) + 2(x^3 - 1)$

Thus my final answer was $(x^3 - 1)(x^2+2)$

From there, I had no clue what to do. My text book says the correct answer is $(x^2+2)(x-1)(x^2+x+1)$ but I just have no clue how they got there.

Any help would be appreciated, thanks.
• Mar 27th 2013, 09:33 PM
MINOANMAN
Re: Factoring a polynomial
Reneg

continue the factorization process. and factorize the (x^3-1) it will give you (x-1)(x^2+x+1). therefore the polynomial after factorization becomes
(x^2+2)(x-1)(x^2+x+1).
now how to factorize the x^3-1 ..the (x-1) is a factor of the polynomial x^3 -1 therefore dividing by (x-1) or simply use Horner's method
(Horner's method - Wikipedia, the free encyclopedia ) you will get the result .

MINOAS
• Mar 30th 2013, 01:37 PM
ReneG
Re: Factoring a polynomial
I just don't see how you can factor out $x - 1$ from $(x^3 - 1)(x^2 + 2)$ because there is no greatest common factor other than 1
• Mar 30th 2013, 01:46 PM
Plato
Re: Factoring a polynomial
Quote:

Originally Posted by ReneG
I just don't see how you can factor out $x - 1$ from $(x^3 - 1)(x^2 + 2)$ because there is no greatest common factor other than 1

$(x^3-1)=(x-1)(x^2+x+1)$ the difference of two cubes.
• Mar 30th 2013, 03:05 PM
HallsofIvy
Re: Factoring a polynomial
Do you know how to multiply polynomials? What do you get when multiply $(x- 1)(x^2+ x+ 1)$?
• Mar 30th 2013, 04:49 PM
ReneG
Re: Factoring a polynomial
Oh wow, completely forgot about that rule. Seeing 1 as a cube root of 1 wasn't really intuitive for me at first, thanks!