How do you simplify #sqrt (1242)#?

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Answer 1 · 2016-04-17T15:40:58

#3sqrt(138)#

First decompose 1242 in prime factors:

It ends in 2 so it is dividible by 2:

#1242/2=621#

The sum of the digits of 621 is 9, therefore it is multiple o #3^2#:

#621/9=69#

#The sum of digits is multiple of 3, so it is divible by 3 again:

#69/3=23#

23 is prime.

#sqrt(1242)=sqrt(2*3^2*3*23)=3sqrt(2*3*23)=3sqrt(138)#