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

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

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- Nov 18th 2006, 09:13 PMputnam120polynomial root
If we are told that $\displaystyle a+\frac{b}{2}+\frac{c}{3}+\frac{d}{5}=-\frac{e}{6}$, where $\displaystyle a,b,c,d,e \in {R}$.

Prove that the polynomial $\displaystyle f(x)=a+bx+cx^2+dx^4+ex^5$ has at least one real zero. - Nov 19th 2006, 12:23 AMCaptainBlack
As:

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

and:

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

$\displaystyle f(x)$ 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 AMThePerfectHacker
Use mean value theorem for integral.

Consider the continous function on $\displaystyle [0,1]$, $\displaystyle f(x)=a+bx+cx^2+dx^4+ex^5$.

By the integral mean value theorem there is a number $\displaystyle c\in [0,1]$ such as,

$\displaystyle f(c)(1-0)=\int_0^1 a+bx+cx^2+dx^4+ex^5dx$

But that tells us that,

$\displaystyle f(c)=ax+\frac{bx^2}{2}+\frac{cx^3}{3}+\frac{dx^5}{ 5}+\frac{ex^6}{6} \big|_0^1$

Evaluate,

$\displaystyle f(c)=a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}+\frac{e }{6}=0$

Note only we know it has a solution we know it has a solution on the interval $\displaystyle [0,1]$ - Nov 19th 2006, 10:20 AMThePerfectHacker
- Nov 19th 2006, 10:31 AMCaptainBlack
Well this partiular statement is independent of $\displaystyle e$ as I said any odd order

polynomial, but in this case yes I suppose $\displaystyle e$ 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

$\displaystyle e=0$ (but I don't intend to check that now).

RonL