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.
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
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 $\displaystyle [0,1]$
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