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

$\left( t+\dfrac{1}{3}\right)^2$
The 2 numbers whose product is $ac= 1\left( \dfrac{1}{9} \right)=\dfrac{1}{9}$ and whose sum is $b= \dfrac{2}{3}$ are $\left\{ \dfrac{1}{3}, \dfrac{1}{3} \right\}.$ Using these two numbers to decompose the middle term of the given trinomial, $t^2+\dfrac{2}{3}t+\dfrac{1}{9} ,$ then the factored form is \begin{array}{l} t^2+\dfrac{1}{3}t+\dfrac{1}{3}t+\dfrac{1}{9} \\\\= \left( t^2+\dfrac{1}{3}t \right) +\left( \dfrac{1}{3}t+\dfrac{1}{9}\right) \\\\= t\left( t+\dfrac{1}{3} \right) +\dfrac{1}{3}\left( t+\dfrac{1}{3}\right) \\\\= \left( t+\dfrac{1}{3}\right)\left( t+\dfrac{1}{3}\right) \\\\= \left( t+\dfrac{1}{3}\right)^2 .\end{array}