1
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$x^3-2x^2-5x+6$

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?

  • 0
    In the Wikipedia example, the remainder is $-123$. Look at that line where they write the result. Divide that entire equation by $(x-3)$.2012-12-07

2 Answers 2

4

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

2

The factor is $x-1$ because $+1$ is a root.