## Intermediate Algebra (6th Edition)

$(3x+14)(x+2)$
Let $z=(x+3)$. Then the given expression, $3(x+3)^2+2(x+3)-5$, is equivalent to $3z^2+2z-5$.\\ The two numbers whose product is $ac= 3(-5)=-15$ and whose sum is $b= 2$ are $\{ 5,-3 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $3z^2+2z-5$, is \begin{array}{l}\require{cancel} 3z^2+5z-3z-5 \\\\= (3z^2+5z)-(3z+5) \\\\= z(3z+5)-(3z+5) \\\\= (3z+5)(z-1) .\end{array} Since $z=(x+3)$, then, \begin{array}{l} (3z+5)(z-1) \\\\= (3(x+3)+5)((x+3)-1) \\\\= (3x+9+5)(x+3-1) \\\\= (3x+14)(x+2) .\end{array}