Find the square root of $\dfrac{8!}{70}$.  (Reminder: The number $n!$ is the product of the integers from 1 to $n$.  For example, $5!=5\cdot 4\cdot3\cdot2\cdot 1= 120$.)
To find the square root of a fraction, I can take the square root of the numerator and the square root of the denominator separately, as long as they are both nonnegative. So, I need to find $\sqrt{\dfrac{8!}{70}}=\dfrac{\sqrt{8!}}{\sqrt{70}}$. I know that $8!=8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$, so I can try to factor it into some perfect squares. I see that $8=2^3$, $6=2\cdot 3$, $4=2^2$, and $2=2$, so I can write $8!=2^7\cdot 3^2\cdot 5\cdot 7$. This means that $\sqrt{8!}=\sqrt{2^7\cdot 3^2\cdot 5\cdot 7}=\sqrt{2^6\cdot 2\cdot 3^2\cdot 5\cdot 7}=2^3\cdot 3\cdot \sqrt{2\cdot 5\cdot 7}$. Now, I need to find $\sqrt{70}$. I know that $70=2\cdot 5\cdot 7$, so $\sqrt{70}=\sqrt{2\cdot 5\cdot 7}$. Therefore, $\sqrt{\dfrac{8!}{70}}=\dfrac{\sqrt{8!}}{\sqrt{70}}=\dfrac{2^3\cdot 3\cdot \sqrt{2\cdot 5\cdot 7}}{\sqrt{2\cdot 5\cdot 7}}$. I can cancel out the common factors of $\sqrt{2\cdot 5\cdot 7}$ in the numerator and denominator, and get $\sqrt{\dfrac{8!}{70}}=2^3\cdot 3=8\cdot 3=24$. 