Answer
$2y^7-12y^6-32y^5$
Work Step by Step
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}
