Factorize polynomials function

Question

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

Answer 1

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.

Answer 2

$$x^4-6x^2-8x+24=(x^2+5)^2-16x^2-8x-1=(x^2+5)^2-(4x+1)^2=$$ $$=(x^2-4x+4)(x^2+4x+6)=(x-2)^2(x^2+4x+6),$$ which gives the answer: $$\{2,-2+\sqrt{2}i,-2-\sqrt2i\}.$$