# What is wrong with my polynomial division? $x^3-2x^2-5x+6$

$$x^3-2x^2-5x+6$$

I want to get the solutions of this. I did a polynomial division.

First, I know that $(x+1)$ is a factor since $1^3-2\cdot1^2-5\cdot1+6 = 0$

So my division goes like this: $$..........x^2-3x-2$$ $$(x+1)|\overline{x^3-2x^2-5x+6}$$ $$x^3+x^2$$ $$......\overline{0-3x^2}-5x$$ $$........-3x^2-3x$$ $$...............\overline{0-2x}+6$$ $$....................2x-2$$ $$....................\overline{0 + 8}$$

(Sorry for the improvised formatting. Ignore any dots you see there.)

So I get, at the end, the quadratic $x^2-3x-2+8=x^2-3x+6$

However, $\triangle = (-3)^2-4(1)(6)=9-4(6)=9-24=-15$

Therefore, there are no solutions since $\triangle$ is negative.

But I definitely did something wrong, since I do know that the solutions are $1,-2,3$

What did I do wrong?

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FYI, your "end quadratic" would actually be $x^2 -3x -2 + \dfrac{8}{x+1}$ (assuming the ill-advised division was carried out correctly.) –  The Chaz 2.0 Dec 7 '12 at 7:41
@TheChaz: Oh, didn't know that at all. Thanks. Wait... could you explain? I mean, en.wikipedia.org/wiki/Polynomial_long_division doesn't seem to do what with the remainder –  Zol Tun Kul Dec 7 '12 at 7:58
In the Wikipedia example, the remainder is $-123$. Look at that line where they write the result. Divide that entire equation by $(x-3)$. –  The Chaz 2.0 Dec 7 '12 at 14:46

$(x+1)$ isn't a factor; you discovered that $1$ is a zero of your polynomial, hence $(x-1)$ is a factor.
The factor is $x-1$ because $+1$ is a root.