How do you simplify #(8sqrt2)/(2sqrt8)#?

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Answer 1 · 2018-03-19T00:21:47

#=>+-2#

We can't factor out a perfect square out of the numerator, so it'll remain the same. Let's see what we can do with the denominator.

We can factor #sqrt8# into #sqrt2*sqrt4#. Thus we have:

#(8cancelsqrt2)/(2cancelsqrt2*sqrt4)#

#sqrt2# obviously cancels with itself, and we get:

#8/(2sqrt4)#

Which can be simplified to

#8/(2*+-2)#

#=>8/(+-4)#

#=>+-2#