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# polynomial equation of odd degree

###### Theorem.

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 $x$.

Keywords:

odd degree, real coefficients

Related:

AlgebraicEquation, ExampleOfSolvingACubicEquation

Type of Math Object:

Theorem

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

26A15*no label found*26A09

*no label found*12D10

*no label found*26C05

*no label found*

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