## Intermediate Algebra (6th Edition)

$4x(3y-2z^2)^2$
Factoring the $GCF=4x$, then the given expression, $36xy^2-48xyz^2+16xz^4$, is equivalent to \begin{array}{l} 4x(9y^2-12yz^2+4z^4) .\end{array} The two numbers whose product is $ac= 9(4)=36$ and whose sum is $b= -12$ are $\{ -6,-6 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $4x(9y^2-12yz^2+4z^4)$, is \begin{array}{l}\require{cancel} 4x(9y^2-6yz^2-6yz^2+4z^4) \\\\= 4x[(9y^2-6yz^2)-(6yz^2-4z^4)] \\\\= 4x[3y(3y-2z^2)-2z^2(3y-2z^2)] \\\\= 4x[(3y-2z^2)(3y-2z^2)] \\\\= 4x(3y-2z^2)^2 .\end{array}