# polynomial root

• Nov 18th 2006, 09:13 PM
putnam120
polynomial root
If we are told that $a+\frac{b}{2}+\frac{c}{3}+\frac{d}{5}=-\frac{e}{6}$, where $a,b,c,d,e \in {R}$.

Prove that the polynomial $f(x)=a+bx+cx^2+dx^4+ex^5$ has at least one real zero.
• Nov 19th 2006, 12:23 AM
CaptainBlack
Quote:

Originally Posted by putnam120
If we are told that $a+\frac{b}{2}+\frac{c}{3}+\frac{d}{5}=-\frac{e}{6}$, where $a,b,c,d,e \in {R}$.

Prove that the polynomial $f(x)=a+bx+cx^2+dx^4+ex^5$ has at least one real zero.

As:

$x \to \infty$ $f(x) \to + \infty$

and:

$x \to -\infty$ $f(x) \to - \infty$

$f(x)$ as its continuous has a real root.

In fact all odd order polynomial with real coefficients must have at
least one real root.

RonL
• Nov 19th 2006, 06:00 AM
ThePerfectHacker
Quote:

Originally Posted by putnam120
If we are told that $a+\frac{b}{2}+\frac{c}{3}+\frac{d}{5}=-\frac{e}{6}$, where $a,b,c,d,e \in {R}$.

Prove that the polynomial $f(x)=a+bx+cx^2+dx^4+ex^5$ has at least one real zero.

Use mean value theorem for integral.
Consider the continous function on $[0,1]$, $f(x)=a+bx+cx^2+dx^4+ex^5$.

By the integral mean value theorem there is a number $c\in [0,1]$ such as,
$f(c)(1-0)=\int_0^1 a+bx+cx^2+dx^4+ex^5dx$
But that tells us that,
$f(c)=ax+\frac{bx^2}{2}+\frac{cx^3}{3}+\frac{dx^5}{ 5}+\frac{ex^6}{6} \big|_0^1$
Evaluate,
$f(c)=a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}+\frac{e }{6}=0$
Note only we know it has a solution we know it has a solution on the interval $[0,1]$
• Nov 19th 2006, 10:20 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank

In fact all odd order polynomial with real coefficients must have at
least one real root.

But how do you know that $e\not = 0$
• Nov 19th 2006, 10:31 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
But how do you know that $e\not = 0$

Well this partiular statement is independent of $e$ as I said any odd order
polynomial, but in this case yes I suppose $e$ could in principle
be zero.

In fact I do have another method using Descartes rule of signs which
does use the condition, which I beleive handles the possibility that
$e=0$ (but I don't intend to check that now).

RonL