How do you simplify #sqrt(72x^3y^4z^5)#?

Question

1950 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-22T17:11:09

# = 6 x y ^2 z^2 sqrt ( 2 x z#

#sqrt ( color(blue)(72) x ^3 y^4 z ^5#

# = sqrt ( color(blue)(3 * 3 * 2 * 2 * 2) * x ^2 * x * y^2 * y^2 * z ^2 * z^2 * z#

# = sqrt ( color(blue)(3 ^2 * 2^2 * 2) * x ^2 * x * y^2 * y^2 * z ^2 * z^2 * z#

# = 6sqrt ( 2 * color(green)(x ^2) * x * y^2 * y^2 * z ^2 * z^2 * z#

# = 6 * x sqrt ( 2 * x * color(green)( y^2 * y^2) * z ^2 * z^2 * z#

# = 6 * x * y ^2 sqrt ( 2 * x * color(green)( z ^2 * z^2) * z#

# = 6 * x * y ^2 * z^2 sqrt ( 2 * x * z#

# = 6 x y ^2 z^2 sqrt ( 2 x z#