<s>[INST] Factorize polynomials function The task was: 
Find the set of zeros (roots) of the following function $$f(x)=x^4-6x^2-8x+24$$
What I did:
I found the possibles roots $$\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24  $$
and I found that X=2 but I cannot get the other 3 roots
 [/INST] You found that $x = 2$ is a root of $f(x) = x^4 - 6x^2 - 8x + 24$.  This means $x - 2$ is a factor.  Dividing $f(x)$ by $x - 2$ yields 
\begin{align*}
f(x) & = x^4 - 6x^2 - 8x + 24\\
     & = x^3(x - 2) + 2x^3 - 6x^2 - 8x + 24\\
     & = x^3(x - 2) + 2x^2(x - 2) + 4x^2 - 6x^2 - 8x + 24\\
     & = x^3(x - 2) + 2x^2(x - 2) - 2x^2 - 8x + 24\\
     & = x^3(x - 2) + 2x^2(x - 2) - 2x(x - 2) - 4x - 8x + 24\\
     & = x^3(x - 2) + 2x^2(x - 2) - 2x(x - 2) - 12x + 24\\
     & = x^3(x - 2) + 2x^2(x - 2) - 2x(x - 2) - 12(x - 2)\\
     & = (x - 2)(x^3 + 2x^2 - 2x - 12)
\end{align*}
Now apply the Rational Roots Theorem and Factor Theorem to 
$$g(x) = x^3 + 2x^2 - 2x - 12$$
to find another root, which may be the same as the first root.
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