Grouping All the Way

Sometimes the bracket left over after grouping factors again. Take ${x}^{3}+2{x}^{2}-9x-18$. The first pair is ${x}^{2}\left(x+2\right)$. For the second pair, take out -9, not 9: $-9x-18$ is $-9\left(x+2\right)$. Taking out the minus is what keeps the brackets matching.

Now take out the common factor: $\left(x+2\right)\left({x}^{2}-9\right)$. But ${x}^{2}-9$ is a difference of squares, so it factors too:

$${x}^{3}+2{x}^{2}-9x-18 = \left(x+2\right)\left(x+3\right)\left(x-3\right)$$

Factoring is finished only when no bracket factors any further. Each factor gives a root, so this cubic is zero at -2, -3 and 3.