What is a generic solution to factor $\alpha x^6 - \beta x^4+\gamma x^2 -\delta$

Question

I'm curious if there exists a generic solution /approach to factor the polynomial $\alpha x^6 - \beta x^4+\gamma x^2 -\delta $ with $\alpha \beta \gamma \delta$ the coefficients of the polynomial.

Answer 1

HINT

Let $y=x^2$

$$x^6 - \alpha x^4+\beta x^2 -\gamma\iff y^3 - \alpha x^2+\beta y -\gamma$$

and solve the cubic formula.

Answer 2

Substitute $t=x^2$ to get $\alpha x^6 - \beta x^4+\gamma x^2 -\delta= \alpha t^3 - \beta t^2+\gamma t -\delta$, then solve the cubic equation using cardano's formula

Answer 3

Set $y=x^2$. Then $y^3-ay^2+by-c=0$ can be easily tested for a linear factor in $y$. Also, one can solve the cubic equation in $y$.

Answer 4

There is sometimes an alternative. The demand that roots come in $\pm$ pairs is pretty restrictive, though.

This first type is well behaved, example

$$ ( x^2 - 2) ( x^2 - 3) ( x^2 - 5 ) = x^6 - 10x^4 + 31x^2 - 30 $$ There are six real roots, square roots of things. The general pattern would be $$ (x^2 - a)(x^2 - b)(x^2 - c) = x^6 - (a+b+c)x^4 + (bc + ca + ab)x^2 - abc.$$ If $a,b,c$ are all positive, real roots.

We could also have $$ (x^2 - a)(x^4 + b x^2 + c) $$ with both factors irreducible over the rationals, say.

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Let $$ x^3 + a x^2 + b x + c $$ have no rational roots, just be really unpleasant. We still get your $$ \left(x^3 + a x^2 + b x + c \right)\left(x^3 - a x^2 + b x - c \right)= x^6 + (2b-a^2) x^4 + (b^2 - 2ac) x^2 - c^2 $$

Here is a good one

$$ \left(x^3 + x^2 -12 x + 11 \right)\left(x^3 - x^2 -12 x - 11 \right)= x^6 -25 x^4 +122x^2 - 121 $$

The first cubic was constructed using the cyclotomic methods of Gauss. Its roots are sums of cosines