Chapter 5 - Polynomials and Factoring - 5.6 Factoring: A General Strategy - 5.6 Exercise Set - Page 344: 9

$(3t+7)(t+3) $

The $2$ numbers whose product is $ac= 3(21)=63 $ and whose sum is $b= 16 $ are ${ 7,9 }.$ Using these $2$ numbers to decompose the middle term of the given expression, $ 3t^2+16t+21 ,$ then the factored form is \begin{array}{l} 3t^2+7t+9t+21 \\\\= (3t^2+7t)+(9t+21) \\\\= t(3t+7)+3(3t+7) \\\\= (3t+7)(t+3) .\end{array}