Question

express in simplest radical form assuming $x > 0$.
$x^{2}\sqrt{x^{8}}$
answer
$\circ$ $x^{10}$ $\circ$ $x^{2}\sqrt{x^{8}}$
$\circ$ $x^{6}$ $\circ$ $x^{6}\sqrt{x}$
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Explanation:

Step1: Simplify the radical term

Recall that $\sqrt{x^n}=x^{\frac{n}{2}}$ for $x>0$. So $\sqrt{x^8}=x^{\frac{8}{2}}=x^4$.

Step2: Multiply with the existing term

Multiply $x^2$ by $x^4$. Using exponent rule $x^a \cdot x^b = x^{a+b}$, we get $x^{2+4}=x^6$.

Answer:

C. $x^6$