Question

how do you do 32y^4z^3 under a sq. root sign?

Answer 1

32 is 4*4*2 so sqrt 32 is 4*sqrt2
4 is 2*2 so sqrt 4 is 2
sqrt y^4=y^2
so: 2*4*y^2*sqrt(2z^3)

Answer 2

sqrt(2z^3) with this part you cannot do anything

Answer 3

oh but you can

Answer 4

my guess is you are supposed to write this in "simplest radical form" which means you cannot have a cube under a square root.

Answer 5

\[\sqrt{32y^4z^3}=\sqrt{4^2\times 2 \times y^2 \times y^2 \times z^2 \times z}\]
\[=\sqrt{4^2}\times \sqrt{y^2}\times \sqrt{y^2}\times \sqrt{z^2} \times\sqrt{2z}\]
\[=4y^2z\sqrt{2z}\]

Answer 6

thanks, still confused a lot but thanks

Answer 7

are you still confused?