## anonymous 5 years ago how do you do 32y^4z^3 under a sq. root sign?

1. anonymous

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)

2. anonymous

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

3. anonymous

oh but you can

4. anonymous

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

5. anonymous

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

6. anonymous

thanks, still confused a lot but thanks

7. anonymous

are you still confused?