Question 57580
{{{w^3+5w^2-w=5}}} Set = to 0
{{{w^3+5w^2-w-5=5-5}}}
{{{w^3+5w^2-w-5=0}}} Factor by grouping
{{{(w^3+5w^2)+(-w-5)=0}}} Factor the GCF in each group
{{{w^2(w+5)-1(w+5)=0}}}  Factor out w+5
{{{(w+5)(w^2-1)=0}}}  Factor {{{w^2-1}}}, it's the difference of squares
{{{(w+5)(w+1)(w-1)=0}}} Set each parenthesis = 0 and solve for w.
w+5=0   and w+1=0  and w-1=0
w+5-5=0-5   and w+1-1=0-1  and w-1+1=0+1
w=-5  and w=-1  and w=1
Some books write the solution set as: w={-5,-1,1}
Happy Calculating!!!