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

$t^2-\dfrac{2}{5}t+\dfrac{1}{25}$
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the given expression, $\left( t-\dfrac{1}{5} \right)^2$, is equivalent to \begin{array}{l}\require{cancel} (t)^2+2(t)\left( -\dfrac{1}{5} \right)+\left( -\dfrac{1}{5} \right)^2 \\\\= t^2-\dfrac{2}{5}t+\dfrac{1}{25} .\end{array}