Answer
$a^2+\dfrac{1}{6}a-\dfrac{1}{3}$
Work Step by Step
Using $(a+b)(c+d)=ac+ad+bc+bd$ or the product of 2 binomials, the product of the given expression, $
\left( a-\dfrac{1}{2} \right)\left( a+\dfrac{2}{3} \right)
,$ is
\begin{array}{l}\require{cancel}
a(a)+a\left(\dfrac{2}{3}\right)-\dfrac{1}{2}(a)-\dfrac{1}{2}\left(\dfrac{2}{3}\right)
\\\\=
a^2+\dfrac{2}{3}a-\dfrac{1}{2}a-\dfrac{1}{3}
\\\\=
a^2+\dfrac{4}{6}a-\dfrac{3}{6}a-\dfrac{1}{3}
\\\\=
a^2+\dfrac{1}{6}a-\dfrac{1}{3}
.\end{array}
