A Quadratic in Disguise

${x}^{4}-5{x}^{2}+4$ has degree 4, but look at its powers: ${x}^{4}$ is ${\left({x}^{2}\right)}^{2}$. Write u for ${x}^{2}$ and it becomes ${u}^{2}-5u+4$, an ordinary quadratic, which factors as $\left(u-1\right)\left(u-4\right)$.

Put ${x}^{2}$ back in for u: $\left({x}^{2}-1\right)\left({x}^{2}-4\right)$. Both brackets are differences of squares, so keep going:

$${x}^{4}-5{x}^{2}+4 = \left(x+1\right)\left(x-1\right)\left(x+2\right)\left(x-2\right)$$

The trick works whenever the powers are ${x}^{4}$, ${x}^{2}$ and a number: the top power is double the middle one, so the middle power plays the part of x.