#### Answer

$(2k^2+1)(k^2-3)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
2k^4-5k^2-3
,$ 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 $
2(-3)=-6
$ and the value of $b$ is $
-5
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-6\}, \{2,-3\},
\\
\{-1,6\}, \{-2,3\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
1,-6
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2k^4+k^2-6k^2-3
.\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}
(2k^4+k^2)-(6k^2+3)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
k^2(2k^2+1)-3(2k^2+1)
.\end{array}
Factoring the $GCF=
(y^2-2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2k^2+1)(k^2-3)
.\end{array}