How should I factor this polynomial: $x^3 - x^2 - 4x - 6$
How to factorize a cubic equation?
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0[Rational root test](http://en.wikipedia.org/wiki/Rational_root_test) – 2012-12-16
3 Answers
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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0Thanks :) 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? – 2012-12-16
Hint: the polynomial evaluates to zero when $x=3$.
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.