How to factorize the polynomial $x^4+3x^2+x$

Question

$$x^4+3x^2+x=0$$

I only can think of

$$x(x^3+3x+1)$$

I don't know what else to do if substitution fails and also RRT.

Is this solvable?

or we have to use imaginary solutions?

There is a cubic formula you can look up to factor $x^3+3x+1$ — Hagamena, Dec 13 '23 at 03:48
A local discussion explaining Cardano's cubic formula. RRT implies that the cubic is irreducible over $\Bbb{Q}$ but obviously it has a single real root, and hence it factors further over the reals. — Jyrki Lahtonen, Dec 13 '23 at 04:00

score 2 · Accepted Answer · answered Dec 13 '23 at 02:53

At a precalculus level, that's as far as you can factorise it. As Wolfram Alpha demonstrates, the cubic component has one real root at

$$x = \sqrt[3]{\frac{\sqrt{5} - 1}{2}} - \sqrt[3]{\frac{2}{\sqrt{5} - 1}}$$

as well as two complex roots, but finding any of those requires a bit of algebraic trickery.

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