Question #2019c

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Answer 1 · 2017-03-03T15:11:27

#x(x+2)(x-2)=0#

so #x=-2,0,2#

You can take out a factor of #x# because everything has at least one #x#, so

#x^3-4x=x(x^2-4)#

Now you have a squared #x# minus a square number, which is always in the form #(x+b)(x-b)# where #b# is the root of the square number.

#x^2-4=(x+2)(x-2)#

So you can say the whole thing is

#x^3-4x=x(x+2)(x-2)#

which gives the solutions of #x=0,-2, +2#