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

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Answer 1 · 2016-03-26T06:22:10

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

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

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

How do you factor completely a5b−aa^5b-a?