# How do you simplify 3sqrt(27x^7)?

Mar 14, 2018

$9 {x}^{2} \sqrt{3 {x}^{3}}$

#### Explanation:

Rule: group things inside the square root in groups of 2 so you can "take them out" of the square root

$\sqrt{27}$ can be written as $\sqrt{3 \cdot 3 \cdot 3}$. Now "take it outside the square root $3 \cdot \sqrt{3}$

Then do the same thing for x^7. Recall the rules for exponents and how they add i.e. ${x}^{2} \cdot {x}^{2}$ is equal to ${x}^{4}$

So ${x}^{7}$ rewritten in groups of 2 would be ${x}^{2} \cdot {x}^{2} \cdot {x}^{3}$. Since we can take those two groups out, it becomes ${x}^{2} \cdot \sqrt{{x}^{3}}$

Now combine $3 \cdot \sqrt{3}$ and ${x}^{2} \cdot \sqrt{{x}^{3}}$ so you get $9 {x}^{2} \sqrt{3 {x}^{3}}$