How do you find all zeros of the function #2x^6-3x^2-x+1#?
By the rational roots theorem, the only possible rational zeros are expressible in the form
That means that the only possible rational zeros are:
If we write
We can find them numerically, for example by using Newton's method.
#f'(x) = 6x^5-6x-1#
If we choose an initial approximation
#a_(i+1) = a_i - f(x)/(f'(x)) = a_i - (x^6-3x^2-x+1)/(6x^5-6x-1)#
The two Real zeros can readily be found by putting these formulae into a spreadsheet. To find the four Complex zeros is a little more complicated. If your spreadsheet application (like mine) does not handle Complex numbers directly, then you need to separate out Real and imaginary parts in separate columns.
If I have sufficient time, I may create such a spreadsheet.