Answer
$(\sqrt{x} + \sqrt{3})^5$ = $x^2\sqrt{x}$ + $5\sqrt{3}x^2$ + $30x\sqrt{x}$ + $30\sqrt{3}x$ + $45\sqrt{x}$ + $9\sqrt{3}$
Work Step by Step
For $(\sqrt{x} + \sqrt{3})^5$, using the Binomial Theorem,
$(\sqrt{x} + \sqrt{3})^5$
= $(\sqrt{x})^5$ + $\frac{5}{1!}(\sqrt{x})^4(\sqrt{3})^1$ + $\frac{5(5-1)}{2!}(\sqrt{x})^3\sqrt{3}^2$ + $\frac{5(5-1)(5-2)}{3!}(\sqrt{x})^2\sqrt{3}^3$ + $\frac{5(5-1)(5-2)(5-3)}{4!}(\sqrt{x})^1\sqrt{3}^4$ + $(\sqrt{3})^5$
= $x^2\sqrt{x}$ + $5\sqrt{3}x^2$ + $30x\sqrt{x}$ + $30\sqrt{3}x$ + $45\sqrt{x}$ + $9\sqrt{3}$
