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

Jul 6, 2016

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

#### Explanation:

$f \left(x\right) = 3 {x}^{4} + 4 {x}^{3} + x - 1$

Use Descartes' Rule of Signs.

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

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

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