## Intermediate Algebra (12th Edition)

$2y^7-12y^6-32y^5$
Using $(a+b)(c+d)=ac+ad+bc+bd$ or the FOIL Method, the product of the binomials in the given expression, $2y^5(y-8)(y+2) ,$ is \begin{array}{l}\require{cancel} 2y^5[y(y)+y(2)-8(y)-8(2)] \\\\= 2y^5[y^2+2y-8y-16] \\\\= 2y^5[y^2-6y-16] .\end{array} Using $a(b+c)=ab+ac$ or the Distributive Property, the product of the expression above is \begin{array}{l}\require{cancel} 2y^5(y^2)+2y^5(-6y)+2y^5(-16) \\\\= 2y^7-12y^6-32y^5 .\end{array}