Answer
$20(x-6)(x-5)$
Work Step by Step
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}
