The Factor Theorem

The graph of $p\left(x\right) = {x}^{3}-2{x}^{2}-x+2$ crosses the x-axis at -1, 1 and 2. And $p\left(x\right) = \left(x+1\right)\left(x-1\right)\left(x-2\right)$. Each crossing at x = a gives a factor $x-a$. A factor is a bracket that divides p with no remainder.

The reason: at a crossing, $p\left(a\right) = 0$. So dividing by $x-a$ leaves remainder 0. This is the factor theorem. To test $x-3$, work out $p\left(3\right) = 8$. It is not 0, so $x-3$ is not a factor.