# How do you simplifiy \sqrt { 1573x ^ { 7} y ^ { 2} z ^ { 4} }?

Apr 9, 2018

$11 | \setminus {x}^{3} y | {z}^{2} \sqrt{13 x}$

#### Explanation:

Let's find the perfect squares in the radical. Recall that $\sqrt{a b} = \sqrt{a \cdot b}$, $\sqrt{{a}^{b + c}} = \sqrt{{a}^{b} \cdot {a}^{c}}$, and ${\left({a}^{b}\right)}^{c} = {a}^{b \cdot c} = {a}^{b c}$

$\sqrt{121 \cdot 13 \cdot {x}^{6} \cdot x \cdot {y}^{2} \cdot {z}^{4}} \rightarrow$ 121, ${x}^{6}$, ${y}^{2}$, and ${z}^{4}$ are all perfect squares. They can be taken out of the radical.

${11}^{2} = 121$

${\left({x}^{3}\right)}^{2} = {x}^{6}$

${y}^{2} = {y}^{2}$

${\left({z}^{2}\right)}^{2} = {z}^{4}$

$11 | \setminus {x}^{3} y | {z}^{2} \sqrt{13 x} \rightarrow$ Absolute value bars need to be placed around the odd exponents so that they are positive (not around the even ones)

Apr 10, 2018

$\sqrt{1573 {x}^{7} {y}^{2} {z}^{4}} = 11 \sqrt{13} {x}^{\frac{7}{2}} | y | {z}^{2}$

#### Explanation:

$\sqrt{1573 {x}^{7} {y}^{2} {z}^{4}} = \sqrt{13} {\left({11}^{2} {x}^{7} {y}^{2} {z}^{4}\right)}^{\frac{1}{2}}$

$= 11 \sqrt{13} {x}^{\frac{7}{2}} | y | {z}^{2}$.