Question 171465


{{{(2y^3-6y^2-8y)/(4y^3-8y^2-32y)}}} Start with the given expression.



{{{(2*y*(y+1)*(y-4))/(4y^3-8y^2-32y)}}} Factor {{{2y^3-6y^2-8y}}} to get {{{2*y*(y+1)*(y-4)}}}.



{{{(2*y*(y+1)*(y-4))/(2*2*y*(y+2)*(y-4))}}} Factor {{{4y^3-8y^2-32y}}} to get {{{2*2*y*(y+2)*(y-4)}}}.



{{{(highlight(2)*highlight(y)*(y+1)highlight((y-4)))/(highlight(2)*2*highlight(y)*(y+2)highlight((y-4)))}}} Highlight the common terms. 



{{{(cross(2)*cross(y)*(y+1)cross((y-4)))/(cross(2)*2*cross(y)*(y+2)cross((y-4)))}}} Cancel out the common terms. 



{{{(y+1)/(2(y+2))}}} Simplify. 



{{{(y+1)/(2y+4)}}} Distribute



So {{{(2y^3-6y^2-8y)/(4y^3-8y^2-32y)}}} simplifies to {{{(y+1)/(2y+4)}}}.