#### Answer

$(3m^4-4)(2m^4-7)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6m^8-29m^4+28
,$ 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(28)=168
$ and the value of $b$ is $
-29
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-8,-21
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
6m^8-8m^4-21m^4+28
.\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}
(6m^8-8m^4)-(21m^4-28)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2m^4(3m^4-4)-7(3m^4-4)
.\end{array}
Factoring the $GCF=
(3m^4-4)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3m^4-4)(2m^4-7)
.\end{array}