How do you determine the number of complex roots of a polynomial of degree n?

1 Answer
Sep 20, 2016

Answer:

See explanation...

Explanation:

Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra (FTOA) tells us that any non-constant polynomial in one variable with Complex (possibly Real) coefficients has a zero in #CC# (the set of Complex numbers).

A straightforward corollary of this (often stated as part of the FTOA) is that a polynomial of degree #n# with Complex (possibly Real) coefficients has exactly #n# Complex (possibly Real) zeros counting multiplicity.

So a simple answer to your question would be that a polynomial of degree #n# has exactly #n# Complex zeros counting multiplicity.

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How many of those #n# zeros are Real and how many non-Real?

If the polynomial has Real coefficients, then any Complex zeros will occur in Complex conjugate pairs. So the number of non-Real zeros will be even.

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Descartes' Rule of Signs

If the coefficients are Real then we can find out some more things about the zeros by looking at the signs of the coefficients.

If #f(x)# is written in standard form with descending powers of #x# then look at the pattern of signs of coefficients. The number of changes gives you the maximum possible number of positive Real zeros. If there are fewer positive Real zeros then it is fewer by an even number.

To determine the possible number of negative Real zeros, look at the signs of the coefficients of #f(-x)#. This is the same as reversing the sign on terms of odd degree.

For example, consider:

#f(x) = x^4+x^3-x^2+x-2#

The signs of the coefficients are in the pattern #+ + - + -#

Since there are #3# changes of sign, there are #3# or #1# positive Real zeros.

#f(-x) = x^4-x^3-x^2-x-2#

has coefficients with signs #+ - - - -#

Since there is #1# change of sign, #f(x)# has exactly #1# negative Real zero.

Since the total number of zeros of #f(x)# is #4#, that means it has #0# or #2# non-Real Complex zeros.

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Discriminants

Discriminants are another useful tool, which I will describe in another answer.