Fourth Powers

${x}^{4}-17{x}^{2}+16 = 0$ looks like a hard equation of degree four, but it only holds ${x}^{4}$ and ${x}^{2}$, and ${x}^{4}$ is ${\left({x}^{2}\right)}^{2}$. Read ${x}^{2}$ as one unknown and it is a quadratic: that unknown squared, minus 17 of it, plus 16. Two numbers that multiply to 16 and add to -17 are -1 and -16, so it factors as $\left({x}^{2}-1\right)\left({x}^{2}-16\right) = 0$.

Each bracket is a difference of squares, so the left side breaks into four factors: $\left(x+1\right)\left(x-1\right)\left(x+4\right)\left(x-4\right)$. Split into cases and you get four answers: 1, -1, 4 and -4. An equation of degree four has at most four solutions, and this one has all of them.

Not every bracket gives real answers. ${x}^{4}-21{x}^{2}-100 = 0$ factors as $\left({x}^{2}-25\right)\left({x}^{2}+4\right) = 0$, and ${x}^{2} = -4$ has no real solution, since the square of a real number is never negative. The only real solutions are 5 and -5.