# How do you simplify sqrt(81x^4y^9z^2)?

Apr 7, 2015

Find the squares under the root and take them out (un-squared)

$\sqrt{81 {x}^{4} {y}^{9} {z}^{2}} = \sqrt{{9}^{2} \cdot {\left({x}^{2}\right)}^{2} \cdot {\left({y}^{4}\right)}^{2} \cdot y \cdot {z}^{2}} =$

$9 {x}^{2} {y}^{4} z \sqrt{y}$

(the odd $y$ remains under the root)

Apr 7, 2015

To the extent possible factor the argument of the square root into squares:

$\sqrt{81 {x}^{4} {y}^{9} {z}^{2}}$

$= \sqrt{{9}^{2} \cdot {\left({x}^{2}\right)}^{2} \cdot {\left({y}^{4}\right)}^{2} \cdot y \cdot {z}^{2}}$

$= 9 {x}^{2} {y}^{4} z \sqrt{y}$

Apr 7, 2015

9${x}^{2}$${y}^{4}$z$\sqrt{y}$

Except ${y}^{9}$ all other terms are perfect squares, take square root of each of them and write outside the radical sign. Split ${y}^{9}$ as y.${y}^{8}$. Square root of ${y}^{8}$ would be ${y}^{4}$. Write this outside the radical sign. The only term remaining inside the radical sign would be y. Hence the answer would be 9${x}^{2}$${y}^{4}$z$\sqrt{y}$