## Intermediate Algebra (6th Edition)

$y^{2}(x+3)(x+1)$
$x^{2}y^{2}+4xy^{2}+3y^{2}$ Factor out greatest common factor $y^{2}$ $=y^{2}(x^{2}+4x+3)$ To factor $(x^{2}+4x+3)$ find the numbers whose product is $3$ (constant term ) and the sum is $4$(middle term). Factors are $3$ and $1$ whose product is $3$ and sum is $4$. $=y^{2}(x^{2}+4x+3)$ $=y^{2}(x^{2}+3x+1x+3)$ Factoring by grouping $=y^{2}((x^{2}+3x)+(1x+3))$ $=y^{2}(x(x+3)+1(x+3))$ $=y^{2}(x+3)(x+1)$