## Intermediate Algebra: Connecting Concepts through Application

$(4t-3)(t-10)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $4t^2-43t+30 ,$ 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 $4(30)=120$ and the value of $b$ is $-43 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -3,-40 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4t^2-3t-40t+30 .\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} (4t^2-3t)-(40t-30) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} t(4t-3)-10(4t-3) .\end{array} Factoring the $GCF= (4t-3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (4t-3)(t-10) .\end{array}