Answer
$\left( t+\dfrac{1}{3}\right)^2$
Work Step by Step
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}
