Answer
$\dfrac{3x}{4(3x+ 5)}, x\ne-\dfrac{5}{3}, 0$.
Work Step by Step
Note that:
$\dfrac{3}{2x}\cdot \dfrac{x^2}{6x+10}=\dfrac{3}{2x}\cdot \dfrac{x^2}{2(3x)+2(5)}$
Factor each expression completely:
$=\dfrac{3}{2x}\cdot \dfrac{x^2}{2(3x+ 5)}$
Cancel common factor $x$:.
$=\dfrac{3x}{4(3x+ 5)}, x\ne-\dfrac{5}{3}, 0$
Hence, the lowest term is:
$=\dfrac{3x}{4(3x+ 5)}, x\ne-\dfrac{5}{3}, 0$.
