How do you simplify #sqrt(xyz)*sqrt(2y^3z^4)#?

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Answer 1 · 2017-10-02T21:48:45

#y^2z^2sqrt(2xz)#

#sqrt(xyz)* sqrt(2y^3z^4)=> sqrt(2xy^4z^5)#

we can see this as:

#sqrt(2 xx x xx y^4 xx z^4xxz#

We can now extract any exact roots:

The #sqrt(y^4) = y^2# since #y^2 xx y^2 = y^(2+2)= y^4#

#y^2sqrt(2xx x xx z^4 xx z)#

Do the same for #z^4#:

#y^2z^2sqrt(2 xx x xx z)#

This is as far as we can go, so:

#y^2z^2sqrt(2xz)#