Answer
$\dfrac{7(2x-1)^{2}(2x+5)}{2(x+3)^{1/2}}$ or $\dfrac{7(2x-1)^{2}(2x+5)}{2\sqrt{x+3}}$
Work Step by Step
$3(2x-1)^{2}(2)(x+3)^{1/2}+(2x-1)^{3}(\frac{1}{2})(x+3)^{-1/2}$
Take out common factor $(2x-1)^{2}(x+3)^{-1/2}$:
$(2x-1)^{2}(x+3)^{-1/2}[3(2)(x+3)+(\frac{1}{2})(2x-1)]=...$
$...=(2x-1)^{2}(x+3)^{-1/2}[6(x+3)+(\frac{1}{2})(2x-1)]=...$
Simplify the expression inside brackets:
$...=(2x-1)^{2}(x+3)^{-1/2}\Big(6x+18+x-\dfrac{1}{2}\Big)=...$
$...=(2x-1)^{2}(x+3)^{-1/2}\Big(7x+\dfrac{35}{2}\Big)=...$
Take out common factor $\dfrac{7}{2}$ from $\Big(7x+\dfrac{35}{2}\Big)$:
$...=\dfrac{7}{2}(2x-1)^{2}(x+3)^{-1/2}(2x+5)=...$
Organize the expression:
$...=\dfrac{7(2x-1)^{2}(2x+5)}{2(x+3)^{1/2}}$ or $\dfrac{7(2x-1)^{2}(2x+5)}{2\sqrt{x+3}}$
