#### Answer

$(3k^2-7)(2k^2+5)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6k^4-k^2-35
,$ 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.
$\bf{\text{Solution Details:}}$
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
6(-35)=-210
$ and the value of $b$ is $
-1
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-14,15
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
6k^4-14k^2+15k^2-35
.\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^4-14k^2)+(15k^2-35 )
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2k^2(3k^2-7)+5 (3k^2-7)
.\end{array}
Factoring the $GCF=
(3k^2-7)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3k^2-7)(2k^2+5)
.\end{array}