Answer
$\frac{7(2x-1)^2(2x+5)}{2\sqrt{x+3}}$
Work Step by Step
Factor out $2x-1$ and $x+3$:
$3(2x-1)^{2}(2)(x+3)^{1/2}+(2x-1)^{3}(\displaystyle \frac{1}{2})(x+3)^{-1/2}=(2x-1)^{2}(x+3)^{-1/2}[6(x+3)+(2x-1)(\frac{1}{2})] =(2x-1)^{2}(x+3)^{-1/2}(6x+18+x-\frac{1}{2})=(2x-1)^{2}(x+3)^{-1/2}(7x+\frac{35}{2})=\frac{(2x-1)^2(7x+35/2)}{\sqrt{x+3}}=\frac{(2x-1)^2(14x+35)}{2\sqrt{x+3}}=\frac{7(2x-1)^2(2x+5)}{2\sqrt{x+3}}$
