Question 650721
i believe the solution is to use descartes rule of signs.
the rational roots test helps to find the rational roots if they're real.
this reference talks about his rule.
it tells the maximum number of positive real roots of an equation and the maximum possible number of negative roots of an equation.
<a href = "http://people.richland.edu/james/lecture/m116/polynomials/zeros.html" target = "_blank">http://people.richland.edu/james/lecture/m116/polynomials/zeros.html</a>
a very simple way to determine if the eqution has real roots and around where the real roots lie is to graph the equation.
where the equation crosses the x-axis is where the real roots are.
there's a whole list of operations to be performed to find the roots.
the following references address the procedures for doing that.
<a href = "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut38_zero1.htm" target = "_blank">http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut38_zero1.htm</a>
<a href = "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut39_zero2.htm" target = "_blank">http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut39_zero2.htm</a>
to solve your immediate problem, if the polynomial is of degree greater than 2 and it has only one zero, that must mean the zero is real because any polynomial has the number of roots that are equal to the degree of the polynomial.
not all of them are real.
at least some of them can be complex.
it can't have only 1 zero.
it can have only 1 real zero.
it can't have only 1 complex zero because complex zeroes, if they exist, come in conjugate pairs.
that last statement may be the answer you are looking for.
complex roots come in conjugate pairs therefore you can't have only 1 complex root to a polynomial.
long way to get there, but your answer lies in here somewhere.
here's a reference on complex roots.
<a href = "http://www.themathpage.com/aprecalc/factor-theorem.htm" target = "_blank">http://www.themathpage.com/aprecalc/factor-theorem.htm</a>
you have to get through the rest of it, but there is a section in there that explains that complex roots always come in conjugate pairs.