How do you simplify # (4 sqrt12) * (3 sqrt20)#?

1 Answer
May 27, 2016

Answer:

#= 48 sqrt ( 15) #

Explanation:

#(4 sqrt 12 ) * ( 3 sqrt 20)#

# = 4 * sqrt 12 * 3 * sqrt 20 #

  • Simplifying both #sqrt12# and #sqrt20# by prime factorisation.

Note: Prime factorisation is expressing a number as a product of its prime factors.

  • #sqrt12 = sqrt (2 * 2 * 3) = sqrt (2^2 * 3 ) = color(green)(2 sqrt3#

  • #sqrt20 = sqrt (2 * 2 * 5) = sqrt (2^2 * 5 ) = color(blue)(2 sqrt5#

# 4 * sqrt 12 * 3 * sqrt 20 = 4 * color(green)(2 sqrt3) * 3 * color(blue)(2 sqrt5 #

#4 * color(blue)(2) * sqrt3 * 3 * color(blue)(2 ) * sqrt5 = 4 * 3 * color(blue)( 2 * 2) * sqrt3 * sqrt5 #

#= 48 sqrt3 * sqrt5 #

#= 48 sqrt ( 3 * 5) #

#= 48 sqrt ( 15) #