## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$(y-12)(y-8)$
The 2 numbers whose product is $ac= 1(96)=96$ and whose sum is $b= -20$ are $\left\{ -12,-8 \right\}.$ Using these two numbers to decompose the middle term of the given trinomial, $y^2-20y+96 ,$ then the factored form is \begin{array}{l} y^2-12y-8y+96 \\\\= (y^2-12y)-(8y-96) \\\\= y(y-12)-8(y-12) \\\\= (y-12)(y-8) .\end{array}