## Intermediate Algebra (12th Edition)

$(2k+7r)^2$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $4k^2+28kr+49r^2 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 4 ,b= 28 ,\text{ and } c= 49 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 4(49)=196$ and whose sum is $b$ are $\left\{ 14,14 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 4k^2+14kr+14kr+49r^2 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (4k^2+14kr)+(14kr+49r^2) \\\\= 2k(2k+7r)+7r(2k+7r) \\\\= (2k+7r)(2k+7r) \\\\= (2k+7r)^2 .\end{array}