# How do you simplify sqrt (1242)?

Apr 17, 2016

$3 \sqrt{138}$

#### Explanation:

First decompose 1242 in prime factors:

It ends in 2 so it is dividible by 2:

$\frac{1242}{2} = 621$

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

$\frac{621}{9} = 69$

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

$\frac{69}{3} = 23$

23 is prime.

$\sqrt{1242} = \sqrt{2 \cdot {3}^{2} \cdot 3 \cdot 23} = 3 \sqrt{2 \cdot 3 \cdot 23} = 3 \sqrt{138}$