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How should I factor this polynomial: $x^3 - x^2 - 4x - 6$

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Rational root test – Martin Sleziak Dec 16 '12 at 7:14
up vote 5 down vote accepted

Typically when you have a polynomial of the form $$f(x) = x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_n$$ where $a_k \in \mathbb{Z}$, to factorize it, it is a good idea to first plug in the values of the divisors of $a_n$ in $f(x)$ with the hope that it will evaluate to $0$. In your case, the polynomial is $$f(x) = x^3 - x^2 - 4x - 6$$ The divisors of $-6$ are $\{\pm1, \pm2,\pm3 \}$. We find that $f(\pm1) \neq 0$, $f(\pm2) \neq 0$ and $f(-3) \neq 0$, while $f(3) = 0$.

Hence, we have $f(x) = (x-3) (x^2 + ax + b)$. We get that $-3b = -6$ and $a-3 = -1$. Hence, we get that $b = 2$ and $a = 2$. Hence, $$f(x) = (x-3)(x^2 + 2x + 2) = (x-3)((x+1)^2 + 1) = (x-3) (x+1-i) (x+1+i)$$

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I dont see how you get a-3 = -1 ? – Ghost Dec 16 '12 at 4:58
I get $-3b = -6$ by comparing the constant term and $a-3 = -1$ by comparing the coefficient of $x^2$. – user17762 Dec 16 '12 at 4:59
@Ghost $i$ is such that $i^2 = -1$. Hence, $$(x+1)^2 + 1 = (x+1)^2 - i^2 = (x+1-i)(x+1+i)$$ since $a^2 - b^2 = (a-b)(a+b)$ – user17762 Dec 16 '12 at 5:42
Thanks :) I really appreciate it . Although, it's not possible for i^2 to be -1 . So how does that help us? Doesn't it just complicate things? – Ghost Dec 16 '12 at 5:49

Hint: the polynomial evaluates to zero when $x=3$.

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Graph the polynomial: it appears to have a zero at $x=3$ (and this appears to be the only zero). So divide your polynomial by $x-3$ (by long division) and see what you are left with.

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