How do you simplify #-3sqrt(252)#?

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Answer 1 · 2015-07-27T12:29:55

#=color(blue)(−18sqrt(7)#

#−3sqrt252#

Prime factorising #252#

#=−3sqrt(2*2*3*3*7)#

#=−3sqrt(2^2 *3^2*7)#
#=−3(2*3)sqrt(7)#

#=color(blue)(−18sqrt(7)#

Answer 2 · 2015-07-27T12:33:35

You check to see if you can write #252# as a product of a perfect square and another number.

Since #252# is not a perfect square itself, you can check to see if you can write it as a product of a perfect square and another number.

To do that, find the prime factors of #252#

#{(252 : 2 = 126), (126 : 2 = 63) :} -> 2^2#

#{(63 : 3= 21), (21 : 3 = 7) :} -> 3^2#

#7 : 7 = 1 -> 7^1#

So, you can write #252# as

#252 = 2^2 * 3^2 * 7 = 4 * 9 * 7 = underbrace(36)_(color(blue)(=6^2)) * 7#

This means that the original expression will be equivalent to

#-3 * sqrt(252) = -3 * sqrt(36 * 7)#

#-3 * sqrt(36) * sqrt(7) = -3 * 6 * sqrt(7) = color(green)(-18sqrt(7))#