Question

√128x^7y^10

Answer 1

\[\sqrt{128x^7y^{10}}=\sqrt{2^7x^7y^{10}}\] is a start

Answer 2

I know my perfect square is 64 but not sure what to do with my odd exponents. It says "Assume all variables represent positive real numbers"

Answer 3

then see what comes outside of the radical
two goes in to seven three times with a remainder of 1, so \(2^3\) comes out and a \(2\) stays in
also for \(x\) it is the same \(x^3\) comes out, \(x\) stays in
two goes in to ten five times with no remainder, \(y^5\) comes out

Answer 4

5th root of(128x^7y^10)
= 5th root of(32*4 * (x^5)(x^2) * (y^10)
= 5th root of((2^5)*4 * (x^5)(x^2) * (y^2)^5)
= 2xy^2 (5th root of(4x^2))

Answer 5

final answer is
\[8x^3y^5\sqrt{2x}\]

Answer 6

Thank you satellite 73, tremendously helpful!!

Answer 7

yw (easy when you see what to do right? works with any radical)