## Intermediate Algebra: Connecting Concepts through Application

$(3m^4-4)(2m^4-7)$
$\bf{\text{Solution Outline:}}$ 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}