## College Algebra (11th Edition)

$2(8z-3)(6z-5)$
Let $a=4z-3.$ Then the given expression, $6(4z-3)^2+7(4z-3)-3 ,$ is equivalent to \begin{array}{l}\require{cancel} 6a^2+7a-3 .\end{array} The two numbers whose product is $ac= 6(-3)=-18$ and whose sum is $b= 7$ are $\{ 9,-2 \}.$ Using these two numbers to decompose the middle term, then the factored form of $6a^2+7a-3$ is \begin{array}{l}\require{cancel} 6a^2+9a-2a-3 \\\\= (6a^2+9a)-(2a+3) \\\\= 3a(2a+3)-(2a+3) \\\\= (2a+3)(3a-1) .\end{array} Since $a=4z-3$, then the expression $(2a+3)(3a-1)$ is equivalent to \begin{array}{l}\require{cancel} (2(4z-3)+3)(3(4z-3)-1) \\\\= (8z-6+3)(12z-9-1) \\\\= (8z-3)(12z-10) \\\\= (8z-3)2(6z-5) \\\\= 2(8z-3)(6z-5) .\end{array}