# polynomial equation of odd degree

###### Theorem.

The equation

 $\displaystyle a_{0}x^{n}+a_{1}x^{n-1}+\cdots+a_{n-1}x+a_{n}=0$ (1)

with odd degree $n$ and real coefficients $a_{i}$ ($a_{0}\neq 0$) has at least one real root $x$.

Proof.  Denote by $f(x)$ the left hand side of (1).  We can write

 $f(x)=a_{0}x^{n}[1+g(x)]$

where  $\displaystyle g(x):=\frac{a_{1}}{x}\!+\cdots\!+\!\frac{a_{n-1}}{x^{n-1}}\!+\!% \frac{a_{n}}{x^{n}}$.  But we have  $\displaystyle\lim_{|x|\to\infty}g(x)=0$  because

 $\lim_{|x|\to\infty}\frac{a_{i}}{x^{i}}=0$

for all  $i=1,\,...,\,n$.  Thus there exists an  $M>0$  such that

 $|g(x)|<1\,\,\mbox{for}\,\,|x|\geqq M.$

Accordingly  $1+g(\pm M)>0$  and

 $\mbox{sign}f(\pm M)=(\mbox{sign}a_{0})(\mbox{sign}(\pm M))^{n}\cdot 1=(\mbox{% sign}a_{0})(\pm 1)$

since $n$ is odd.  Therefore the real polynomial function $f$ has opposite signs in the end points of the interval$[-M,\,M]$.  Thus the continuity of $f$ guarantees, according to Bolzano’s theorem, at least one zero $x$ of $f$ in that interval.  So (1) has at least one real root $x$.

Title polynomial equation of odd degree PolynomialEquationOfOddDegree 2013-03-22 15:39:19 2013-03-22 15:39:19 pahio (2872) pahio (2872) 7 pahio (2872) Theorem msc 26A15 msc 26A09 msc 12D10 msc 26C05 AlgebraicEquation ExampleOfSolvingACubicEquation