College Algebra (11th Edition)

$16y^2-8y+1+8yz-2z+z^2$
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}