Question

What are the zero(s) for `f(x) = 2x^6 +x^3+3`?

Answer 1

`f(x)` has six Complex zeros which we can find by recognising that `f(x)` is a quadratic in `x^3`.

Explanation:

`f(x) = 2x^6+x^3+3 = 2(x^3)^2 + x^3+3`

Using the quadratic formula we find:

`x^3 = (-1+-sqrt(1^2-4xx2xx3))/(2*2)`

`=(-1+-sqrt(-23))/4 = (-1+-i sqrt(23))/4`

So `f(x)` has zeros:

`x_(1,2) = root(3)((-1+-i sqrt(23))/4)`

`x_(3,4) = omega root(3)((-1+-i sqrt(23))/4)`

`x_(5,6) = omega^2 root(3)((-1+-i sqrt(23))/4)`

where `omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of unity.