#### Answer

$(y^2-2)(y^2+4)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
y^4+2y^2-8
,$ 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 $
1(-8)=-8
$ and the value of $b$ is $
2
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-8\}, \{2,-4\},
\\
\{-1,8\}, \{-2,4\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-2,4
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
y^4-2y^2+4y^2-8
.\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}
(y^4-2y^2)+(4y^2-8)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
y^2(y^2-2)+4(y^2-2)
.\end{array}
Factoring the $GCF=
(y^2-2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(y^2-2)(y^2+4)
.\end{array}