## Intermediate Algebra (6th Edition)

$y(x-1)(3x-8)$
Factoring the $GCF=y$, then the given expression, $3x^2y-11xy+8y$, is equivalent to \begin{array}{l} y(3x^2-11x+8) .\end{array} The two numbers whose product is $ac= 3(8)=24$ and whose sum is $b= -11$ are $\{ -3,-8 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $y(3x^2-11x+8)$, is \begin{array}{l}\require{cancel} y(3x^2-3x-8x+8) \\\\= y[(3x^2-3x)-(8x-8)] \\\\= y[3x(x-1)-8(x-1)] \\\\= y[(x-1)(3x-8)] \\\\= y(x-1)(3x-8) .\end{array}