1. ## Factoring a polynomial

I need help factoring $\displaystyle x^5+2x^3-x^2-2$

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

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

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

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

Any help would be appreciated, thanks.

2. ## 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

3. ## Re: Factoring a polynomial

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

4. ## Re: Factoring a polynomial

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

$\displaystyle (x^3-1)=(x-1)(x^2+x+1)$ the difference of two cubes.

5. ## Re: Factoring a polynomial

Do you know how to multiply polynomials? What do you get when multiply $\displaystyle (x- 1)(x^2+ x+ 1)$?

6. ## 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!