Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 2

$xy^4(x-2y+3x^2y^2)$

The $GCF$ of the terms is $ xy^4 $ since it is the highest expression that can evenly divide (no remainder) all the given constants. Factoring the $GCF,$ the given expression is equivalent to \begin{array}{l}\require{cancel} x^2y^4-2xy^5+3x^3y^6 \\\\= xy^4(x-2y+3x^2y^2) .\end{array}