Help in factorization

Question

So I was thinking that how can I factorize $x^4-2x^2-8$, or any $4$ degree polynomial, without going for the first factor by trial and error method?

Answer 1

Let $x^2=y$.

Then you have to factorize $y^2-4y-8=P(y)$

$P(y)=(y-2- \sqrt{12})(y-2+ \sqrt{12})$

Thus $$x^4-4x^2-8=(x^2-2- \sqrt{12})(x^2-2+ \sqrt{12})=$$ $$(x+ \sqrt{2+ \sqrt{12}})(x-\sqrt{2+ \sqrt{12}})((x^2-2+ \sqrt{12})$$

Answer 2

$$x^4-2x^2-8=x^4-2x^2+1-3^2=(x^2-4)(x^2+2)=(x-2)(x+2)(x^2+2).$$

We can make factorization for all $4$ degree polynomial over $\mathbb R$ by the Ferrari's method.

Answer 3

Express the function as a quadratic in $x^2$

$(x^2)^2 - 2x^2 -8$

Factor it for $x^2$

$(x^2+2)(x^2-4)$

Write 4 as $2^2$ and expand it,

$(x^2+2)(x+2)(x-2)$