## Intermediate Algebra (6th Edition)

$y^2(y-2)(3y+5)$
Factoring the $GCF=y^2$ of the given expression, $3y^4-y^3-10y^2$, results to \begin{array}{l}\require{cancel} y^2(3y^2-y-10) .\end{array} The two numbers whose product is $ac= 3(-10)=-30$ and whose sum is $b= -1$ are $\{ -6,5 \}$. Using these two numbers to decompose the middle term, then the factored form of the resulting expression, $y^2(3y^2-y-10)$,is \begin{array}{l}\require{cancel} y^2(3y^2-6y+5y-10) \\\\= y^2[(3y^2-6y)+(5y-10)] \\\\= y^2[3y(y-2)+5(y-2)] \\\\= y^2[(y-2)(3y+5)] \\\\= y^2(y-2)(3y+5) .\end{array}