## Intermediate Algebra (6th Edition)

$4(x+3)(x-1)$
Factoring the $GCF= 4$ results to $4(x^2+2x-3)$. The two numbers whose product is $-3$ and whose sum is $2$ are $\{ 3,-1 \}$. Using these numbers to decompose the middle term of the trinomial results to \begin{array}{l} 4(x^2+2x-3) \\= 4(x^2+3x-x-3) \\= 4[(x^2+3x)-(x+3)] \\= 4[x(x+3)-(x+3)] \\= 4[(x+3)(x-1)] \\= 4(x+3)(x-1) .\end{array}