# How do you factor completely a^5b-a?

$\left({a}^{5} b - a\right) = a \left({a}^{4} b - 1\right)$
From the given , $\left({a}^{5} b - a\right)$, there is only one common monomial factor which is the $a$
So $\left({a}^{5} b - a\right) = a \left({a}^{4} b - 1\right)$