Grouping to Solve

When a cubic has four terms and no factor shared by all of them, try grouping. Take ${x}^{3}-3{x}^{2}-9x+27 = 0$. The first two terms share ${x}^{2}$ and the last two share -9, so the left side is ${x}^{2}\left(x-3\right)-9\left(x-3\right)$. Now x - 3 is a common factor: $\left(x-3\right)\left({x}^{2}-9\right) = 0$. The bracket ${x}^{2}-9$ is a difference of squares, so the whole equation is $\left(x-3\right)\left(x+3\right)\left(x-3\right) = 0$.

Splitting into cases gives 3, then -3, then 3 again. The factor x - 3 appears twice, so 3 is a repeated root. In the answer it counts once: the solutions are 3 and -3. An answer is a set of numbers, and a set holds each number once, however many factors point to it. On a graph, a root repeated an even number of times, like 3 here, is where the curve touches the x-axis and turns back. A root repeated an odd number of times crosses: ${\left(x-1\right)}^{3}$ crosses at 1.