Answer
$\color{blue}{(2x-1)(3x-7)}$
Work Step by Step
RECALL:
A quadratic trinomial $ax^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = ac$ and $d+e =b$
If such integers exist, then $ax^2+bx + c = ax^2+dx + ex + c$, and may be factored by grouping.
The given trinomial h as $a=6$, $b=-17$, and $c=7$.
Thus, $ac = 6(7) = 42$
Note that $42=-3(-14)$ and $-3+(-14)=-17$
This means that $d=-3$ and $e=-14$.
Rewrite the middle term of the trinomial as $-3x$ +$(-14x)$ to obtain:
$6x^2-17x+7 =6x^2-3x-14x+7 $
Group the first two terms together and the last two terms together.
Then, factor out the GCF in each group to obtain:
$=(6x^2-3x)+(-14x+7)
\\=3x(2x-1) +(-7)(2x-1)$
Factor out the GCF $2x-1$ to obtain:
$=(2x-1)(3x+(-7))
\\=\color{blue}{(2x-1)(3x-7)}$
