# How do you simplify sqrt(108x^5y^8)?

Feb 22, 2017

#### Answer:

$6 {x}^{2} {y}^{4} \sqrt{3 x}$

#### Explanation:

We can think of sqrt(108x^5y^8 as the same as sqrt(108 xx x^5 xx y^8. Then we can can square root each number individually to get $6 {x}^{2} {y}^{4} \sqrt{3 x}$

So the square root of 108 is $6 \sqrt{3}$ because $\sqrt{108}$ can also be written as $\sqrt{36 \times 3}$. We know the square root of 36 to be 6, so we write 6 outside of the square root. Thus we have $6 \sqrt{3}$ so far.

Then we can do the same for ${x}^{5} \times {y}^{8}$. However, when square rooting powers, we divide the power by the root. For example $\sqrt{{y}^{8}}$ becomes ${y}^{4}$. For $\sqrt{{x}^{5}}$, we get ${x}^{2} \sqrt{x}$ because we can only square root 4 so we have a remainder.

We can then put all of these answers together to get $6 {x}^{2} {y}^{4} \sqrt{3 x}$

Feb 24, 2017

#### Answer:

$6 {x}^{2} {y}^{4} \sqrt{3}$

#### Explanation:

Looking for squared values we have:

$\sqrt{{6}^{2} \times 3 \times {x}^{2} \times {x}^{2} \times x \times {y}^{2} \times {y}^{2} \times {y}^{2} \times {y}^{2}}$

$6 {x}^{2} {y}^{4} \sqrt{3}$

Have a look at: https://socratic.org/s/aCtLpnUs