# Solve this simple polynomial

$$\text{Problem: }{x^3-x^2-x-2 \over x^2 + x - 6}$$ My textbook was able to come up with $(x-2)(x^2+x+1)$ $$\text{Textbook: }{(x-2)(x^2+x+1) \over (x-2)(x+3)}$$

I've tried grouping and using the first and last 3 terms as trinomial but it doesn't work. How do I get that solution? Thank you in advance.

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So what's the question? Put the fraction in least terms? – Ian Coley Apr 3 '14 at 18:44
@IanColey well, if I got (x-2). I could eliminate them. I'm trying to solve a limit problem – Mouse Hello Apr 3 '14 at 18:45
Which limit?$\,$ – Ian Coley Apr 3 '14 at 18:46
Question unclear – Jason Zimba Apr 3 '14 at 18:46
Have you had polynomial or synthetic division? (That should have come up somewhere in pre-calculus.) Dividing $\ x-2 \$ into $\ x^3 - x^2 - x + 2 \$ does give a quotient of $\ x^2 + x + 1 \$ ( or using "2" in synthetic division on $\ 1 \ -1 \ -1 \ -2 \$ produces $\ 1 \ 1 \ 1 \ 0 \$ ) . – RecklessReckoner Apr 3 '14 at 19:28

For the cubic polynomial, by checking the value of it at some integer values of $x$ like $x = 2$ then $2^3 - 2^2 - 2 - 2 = 0$. This means $x - 2$ is a factor, and using long division you get the other factor. For the quadratic expression, you also have $2$ is a root and also $-3$ is also a root. Then it can be factored as $(x - 2)(x + 3)$.
What do you mean by long division? Do you mean $x^3-x^2-x-2 \over (x-2)$ – Mouse Hello Apr 3 '14 at 19:04