How do you find the zeroes of the 1/2x^3-x? I factoried it so it would be 1/2x(x^2)-x. The one zero would then be zero, which matches with my book's answer. But I have no idea how my book got +/- the sq. root of two for the other one. Can you demonstrate, please?
The way you factored makes no sense to me>
Your factors should be
x((1/2)x^2 - 1)
so x = 0 or
(1/2)x^2 - 1 = 0
x^2 = 2
then x = ±√2