Answer
$16y^2-8y+1+8yz-2z+z^2$
Work Step by Step
Using $(a\pm b)^2=a^2\pm2ab+b^2$ or the square of a binomial, the product of the given expression, $
[(4y-1)+z]^2
,$ is
\begin{array}{l}\require{cancel}
(4y-1)^2+2(4y-1)(z)+(z)^2
\\\\=
(4y-1)^2+8yz-2z+z^2
\\\\=
(4y)^2-2(4y)(1)+(1)^2+8yz-2z+z^2
\\\\=
16y^2-8y+1+8yz-2z+z^2
.\end{array}
