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

1 Answer
Jul 6, 2016

Answer:

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.