Answer
$(3k-2)(2k+5)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6k^2+11k-10
,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
To factor the trinomial expression above, note that the value of $ac$ is $
6(-10)=-60
$ and the value of $b$ is $
11
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-60\}, \{2,-30\}, \{3,-20\}, \{4,-15\}, \{5,-12\}, \{6,-10\},
\\
\{-1,60\}, \{-2,30\}, \{-3,20\}, \{-4,15\}, \{-5,12\}, \{-6,10\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-4,15
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
6k^2-4k+15k-10
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(6k^2-4k)+(15k-10)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2k(3k-2)+5(3k-2)
.\end{array}
Factoring the $GCF=
(3k-2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3k-2)(2k+5)
.\end{array}