## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$(7t+3)^2$
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 49t^2+42t+9 \end{array} has $ac= 49(9)=441$ and $b= 42 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 21,21 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 49t^2+21t+21t+9 .\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} (49t^2+21t)+(21t+9) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 7t(7t+3)+3(7t+3) .\end{array} Factoring the $GCF= (7t+3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (7t+3)(7t+3) \\\\= (7t+3)^2 .\end{array}