Question

How many real and complex roots does `3x^4+4x^3+x-1 = 0` have?

Answer 1

It has one positive Real root, one negative Real root and a Complex conjugate pair of non-Real roots.

Explanation:

`f(x) = 3x^4+4x^3+x-1`

Use Descartes' Rule of Signs.

The signs of the coefficients of `f(x)` have the pattern `+ + + -`. With one change of sign, this means that `f(x)` has exactly one positive Real zero.

The signs of the coefficients of `f(-x)` have the pattern `+ - - -`. With one change of sign, this means that `f(x)` has exactly one negative Real zero.

Using the Fundamental Theorem of Algebra we know that `f(x)` has exactly as many zeros as its degree `4`, counting multiplicity. So it must have two non-Real Complex zeros. Since the coefficients of `f(x)` are Real, these Complex zeros must occur as a Complex conjugate pair.