How do you simplify #sqrt(20y^5)#?
We know we can only take perfect squares out of the root, so we have to find the square factors in 20 and in
The latter is easy, we know that
20 is more complicated, but not hard. We just need to factor it, i.e.: see every prime integer divisor it has in order.
20 | 2
10 | 2
5 | 5
1 | /
So we know we can take a 2 out.