# How do you factor completely a(a+1)(a+2) - 3a(a+1)?

We note that $a \left(a + 1\right)$ is common to both parts of the expression, i.e.
$a \left(a + 1\right) \left[\left(a + 2\right) - 3\right] = a \left(a + 1\right) \left(a - 1\right)$
$\left(a - 1\right) a \left(a + 1\right)$