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- Feb 14, 2012

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- Feb 14, 2012

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- Jan 26, 2012

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We rewrite the polynomial as

$$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$

where we are setting it equal to zero. We evaluate $f(1)$ and $f(-1)$:

\begin{align*}

f(1)&=5t+s+3r+q-r-t=4t+2r+s+q\\

f(-1)&=5t-s+3r-q-r-t=4t+2r-s-q.

\end{align*}

Idea: if $f(1)\cdot f(-1)<0$, then by the Intermediate Value Theorem, we would have shown there is a root in $[-1,1]$. Now

$$f(1)\cdot f(-1)=(4t+2r)^{2}-(s+q)^{2}.$$

Not seeing where to go with this. There's nothing stopping $s=q=0$, with $4t+2r\not=0$, in which case I have not proved what I want to prove.

Consider the polynomial \(\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)\).

Then \(\displaystyle p(1)=p(-1)=0\) and so \(\displaystyle p\) has an extremum in \(\displaystyle (-1,1)\), so \(\displaystyle p'(x)\) has a root in \(\displaystyle (-1,1)\) ...

.

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- Feb 14, 2012

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WoW!!! What an elegant way to approach this problem! Well done,Consider the polynomial \(\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)\).

Then \(\displaystyle p(1)=p(-1)=0\) and so \(\displaystyle p\) has an extremum in \(\displaystyle [-1,1]\), so \(\displaystyle p'(x)\) has a root in \(\displaystyle [-1,1]\) ...

.

And there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait for the inputs from other members for now...

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Hint:

\(\displaystyle \int_{-1}^1 p(x) dx=0\)