Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 19

$4(x+3)(x-1)$

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}