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

Question

1495 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-26T06:22:10

$(a^5 b-a)=a(a^4b-1)$

From the given , $(a^5 b-a)$ , there is only one common monomial factor which is the $a$

So $(a^5 b-a)=a(a^4b-1)$

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