Question 102704
{{{b^3m^3-14b^2m^2+49bm}}}
First you will factor out any commons, they each have bm, so we will factor that out;
{{{bm(b^2-14b+49)}}}
Now factor the equation;
bm(bm-7)(bm-7)
{{{bm(bm-7)^2}}}
Hope this helps
:)