How do you simplify #4sqrt8*sqrt2#?

2 Answers
Sep 7, 2017

Answer:

1) #4sqrt(4xx2)*sqrt2#

2) #(4sqrt4*sqrt2#) ⋅ #sqrt2#

3) #4xx2sqrt2*sqrt2#

4) #8*2#

5) #16#

Explanation:

1) Find factors of #8# under the radical.
2) Simplify #sqrt4# to #2# and multiply by the #4# outside
3) Multiply #8sqrt2*sqrt2" "(larrsqrt2*sqrt2# equals 2)
4) #8*2#
5) #16#

Sep 7, 2017

Answer:

#=16#

Explanation:

The two square roots can be combined into one. In this case this is a good idea because the product will be a prefect square.

#4sqrt8 xx sqrt2 = 4sqrt(8xx2) #

#= 4 sqrt(16)#

#= 4xx4#

#=16#