Question 708860
{{{y^3+8y^2-y-8}}} Start with the given expression



{{{(y^3+8y^2)+(-y-8)}}} Group the terms in two pairs



{{{y^2(y+8)-1(y+8)}}} Factor out the GCF {{{y^2}}} out of the first group. Factor out the GCF {{{-1}}} out of the second group



{{{(y^2-1)(y+8)}}} Since we have the common term {{{y+8}}}, we can combine like terms



{{{(y-1)(y+1)(y+8)}}} Factor {{{y^2-1}}} to get {{{(y-1)(y+1)}}} using the difference of squares rule



So {{{y^3+8y^2-y-8}}} factors to {{{(y-1)(y+1)(y+8)}}}



In other words, {{{y^3+8y^2-y-8=(y-1)(y+1)(y+8)}}} for all values of y