## Intermediate Algebra (6th Edition)

$20(x-6)(x-5)$
Factoring the $GCF= 20$ results to $20(x^2-11x+30)$. The two numbers whose product is $30$ and whose sum is $-11$ are $\{ -6,-5 \}$. Using these numbers to decompose the middle term of the trinomial results to \begin{array}{l} 20(x^2-11x+30) \\= 20(x^2-6x-5x+30) \\= 20[(x^2-6x)-(5x-30)] \\= 20[x(x-6)-5(x-6)] \\= 20[(x-6)(x-5)] \\= 20(x-6)(x-5) .\end{array}