# Chapter 5 - Polynomials and Factoring - 5.6 Factoring: A General Strategy - 5.6 Exercise Set: 68

$(2y+3)(6y-5)$

#### Work Step by Step

The given expression, $8y-15+12y^2 ,$ can be re-written as \begin{array}{l} 12y^2+8y-15 .\end{array} The two numbers whose product is $ac= 12(-15)=-180$ and whose sum is $b= 8$ are $\{ 18,-10 \}$. Using these two numbers to decompose the middle term of the expression, $8t^2+14t-15 ,$ then the factored form is \begin{array}{l} 12y^2+18y-10y-15 \\\\= (12y^2+18y)-(10y+15) \\\\= 6y(2y+3)-5(2y+3) \\\\= (2y+3)(6y-5) .\end{array}

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