Question

How do you find all zeros of the function `2x^6-3x^2-x+1`?

Answer 1

See explanation...

Explanation:

By the rational roots theorem, the only possible rational zeros are expressible in the form `p/q` with integers `p` and `q` with `p` a divisor of the constant term `1` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2`, `+-1`

If we write `f(x) = x^6-3x^2-x+1` and evaluate `f(x)` for these values, we find that none work. So `f(x)` has no rational zeros.

Since `f(x)` has degree `6` it is not surprising to find that its roots have no simple closed algebraic formulation.

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_0` for a zero of `f(x)`, then we can find better approximations by iterating using the formula:

`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.