Question 277230
{{{112v^3+34v^2-6v}}} first find what is in common with all 3 variables, each one has atleast 1v and can be divided by a 2 evenly. so if we factor 2v out it should look like this:
{{{2v(56v^2+17v-3)}}} i would split the 17v into 14v+3v
{{{2v(56v^2+14v+3v-3)}}} no u can group them
{{{2v(56v^2+14v)+2v(3v-3)}}} factor out a 14 and 1v in the first group and a 3 in the second group
{{{28v^2(4v+1)+6v(v-1)}}} thats as factored as i can get it
<p>
{{{90s^2-195sw+100w^2}}} highest divisable number is 5 that i can see at first glance so i will factor out a 5
{{{5(18s^2-39sw+20w^2)}}} i would split the -39sw to -9sw+(-30sw)
{{{5(18s^2-9sw+(-30sw)+20w^2)}}} now u can split the 2
{{{5(18s^2-9sw)+5(-30sw+20w^2)}}}
{{{5(18s^2-9sw)+5(-30sw+20w^2)}}} in the first group a 9 and 1s can be factores out, in the second group a 10 and 1w can be factored out
{{{45s(2s-9w)+50w(-3s+2w)}}} thats the most i can factor it
<p>
{{{p^2(x-r)-64(x-r)}}} facter the (x-r)out
{{{(x-r)(p^2-64)}}} theres a perfect squre left behind so factor it out
{{{(x-r)((p-8)(p+8))}}}