## Precalculus (6th Edition)

$(4xy^3)(xy^2-2)$
The student was wrong because he was not ave to completely factor the given polynomial. In his answer, $2xy^3(2xy^2-4)$, the terms of the binomial factor have a common factor of $2$. Factoring out the $2$ gives: $=(2xy^3)(2)(xy^2-2) \\=(4xy^3)(xy^2-2)$