Question

simplify.
$3x^2\sqrt{72x}-\sqrt{8x^5}$
assume that the variable represents a positive real number.

Explanation:

Step1: Simplify $\sqrt{72x}$

Factor 72 into $36\times2$, so $\sqrt{72x}=\sqrt{36\times2x}=6\sqrt{2x}$ (since $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ for $a,b\geq0$ and $\sqrt{36}=6$). Then $3x^2\sqrt{72x}=3x^2\times6\sqrt{2x}=18x^2\sqrt{2x}$.

Step2: Simplify $\sqrt{8x^5}$

Factor $8$ into $4\times2$ and $x^5$ into $x^4\times x$. So $\sqrt{8x^5}=\sqrt{4\times2\times x^4\times x}=\sqrt{4}\cdot\sqrt{x^4}\cdot\sqrt{2x}=2x^2\sqrt{2x}$ (since $\sqrt{x^4}=x^2$ for $x>0$).

Step3: Subtract the two simplified terms

Now we have $18x^2\sqrt{2x}-2x^2\sqrt{2x}$. Combine like terms: $(18 - 2)x^2\sqrt{2x}=16x^2\sqrt{2x}$.

Answer:

$16x^2\sqrt{2x}$