How do you solve #8x^3 - 32x=0#?

1 Answer
A08
Mar 12, 2016

There is no need to use 'Comparing method' as suggested.
#x=0,-2,2#

Explanation:

Given
#8x^3-32x=0#
Inspection reveals that #8x# is a factor of right hand side cubic.
#:. 8x(x^2-4)=0#
The quadratic term can be rewritten using the formula, since #4=2^2#
#a^2-b^2=(a+b)(a-b)#
Hence the factors are
#8x(x+2)(x-2)=0#
Now #8!=0#, setting remaining factors equal to zero, we obtain
#x=0, (x+2)=0 and (x-2)=0#
We obtain three values of #x=0,-2,2#

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