Answer
$(b-3)(6b+1)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6b^2-17b-3
,$ use the factoring of trinomials in the form $ax^2+bx+c.$
$\bf{\text{Solution Details:}}$
In the trinomial expression above, $a=
6
,b=
-17
,\text{ and } c=
-3
.$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac=
6(-3)=-18
$ and whose sum is $b$ are $\left\{
-18,1
\right\}.$ Using these two numbers to decompose the middle term results to
\begin{array}{l}\require{cancel}
6b^2-18b+b-3
.\end{array}
Using factoring by grouping, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(6b^2-18b)+(b-3)
\\\\=
6b(b-3)+(b-3)
\\\\=
(b-3)(6b+1)
.\end{array}