# How do you simplify sqrt(75a^12b^3c^5)?

Feb 4, 2015

We'll need a few properties.

1. First of all, let's recall that $\setminus \sqrt{a \setminus \cdot b} = \setminus \sqrt{a} \setminus \sqrt{b}$. This of course applies also to products of more than two factors.
2. I hope you will be ok with the fact that the square root of a number is that number to the power of $\frac{1}{2}$. If not, tell me in the comments and I will explain this exercise in another way.
3. The third properties is that ${a}^{b + c} = {a}^{b} \setminus \cdot {a}^{c}$.

If this things are ok, for the first point we can separate the roots:
$\setminus \sqrt{75 {a}^{12} {b}^{3} {c}^{5}} = \setminus \sqrt{75} \setminus \sqrt{{a}^{12}} \setminus \sqrt{{b}^{3}} \setminus \sqrt{{c}^{5}}$

Factoring $75$ in prime numbers, we have $\setminus \sqrt{75} = \setminus \sqrt{3 \setminus \cdot {5}^{2}} = \setminus \sqrt{3} \setminus \sqrt{{5}^{2}} = 5 \setminus \sqrt{3}$.

For $\setminus \sqrt{{a}^{12}}$, we have that $\setminus \sqrt{{a}^{12}} = {a}^{\frac{12}{2}} = {a}^{6}$.

For $\setminus \sqrt{{b}^{3}}$, we have that $\setminus \sqrt{{b}^{3}} = {b}^{\frac{3}{2}} = {b}^{1 + \frac{1}{2}} = b \setminus \sqrt{b}$

For $\setminus \sqrt{{c}^{5}}$, we have that $\setminus \sqrt{{c}^{5}} = {c}^{\frac{5}{2}} = {c}^{2 + \frac{1}{2}} = {c}^{2} \setminus \sqrt{c}$.

Putting all the pieces together, we have that
$\setminus \sqrt{75 {a}^{12} {b}^{3} {c}^{5}} = 5 {a}^{6} b {c}^{2} \setminus \sqrt{3 b c}$