# How do you simplify -3x^2y^3zsqrt(75x^9y^4z^10)?

Jun 1, 2015

Remembering one exponential law which states that ${a}^{\frac{n}{m}} = \sqrt[m]{{a}^{n}}$, we can rewrite the function as

$- 3 {x}^{2} {y}^{3} z {75}^{\frac{1}{2}} {x}^{\frac{9}{2}} {y}^{\frac{4}{2}} {z}^{\frac{10}{2}}$

(As this is a square root, we have that, for the exponential rule, $m = 2$)

Now, following another law of exponentials, we have that ${a}^{n} \cdot {a}^{m} = {a}^{n + m}$

$- 3 {x}^{\left(2 + \frac{9}{2}\right)} {y}^{\left(3 + 2\right)} {z}^{\left(1 + \frac{10}{2}\right)} {75}^{\frac{1}{2}}$

$- 3 {x}^{\frac{13}{2}} {y}^{5} {z}^{6} {75}^{\frac{1}{2}}$

We can rewrite the element ${75}^{\frac{1}{2}}$ as ${\left(3 \cdot 25\right)}^{\frac{1}{2}} = {3}^{\frac{1}{2}} \cdot {25}^{\frac{1}{2}}$

Thus, we can combine the first $- 3$ with the just-factored $3$:

$- {\left(3\right)}^{1} \cdot \left({3}^{\frac{1}{2}}\right) {x}^{\frac{13}{2}} {y}^{5} {z}^{6} {75}^{\frac{1}{2}} {25}^{\frac{1}{2}}$

As ${25}^{\frac{1}{2}} = \sqrt{25} = 5$,

$5 {\left(- 3\right)}^{1.5} {x}^{6.5} {y}^{5} {z}^{6}$