Question about finding roots of a polynomial and studying the nature

The number of real roots of the equation $1+frac{x}{1}+frac{x^{2}}{2}+frac{x^{2}}{3}+cdots+frac{x^{7}}{7}=0$
(without factorial) is

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My work

Let,$mathrm{f}(mathrm{x})=1+frac{x}{1}+frac{x^{2}}{2}+frac{x^{3}}{3}+cdots+frac{x^{6}}{6}$

[Let, f has a minimum at $x=x_{0},$ where then $f^{prime}left(x_{0}right)=0$]

$Rightarrow 1+x_{0}+x_{0}^{2}+x_{0}^{3}+x_{0}^{4}+x_{0}^{5}=0$

$Rightarrow frac{x_{0}^{6}-1}{x_{0}-1}=0$

$Rightarrow frac{left(x_{0}^{3}-1right)left(x_{0}^{3}+1right)}{x_{0}-1}=0$

$Rightarrowleft(x_{0}^{2}+x_{0}+1right)left(x_{0}^{2}-x_{0}+1right)left(x_{0}+1right)=0$

Which has a real root $x_{0}=-1$

But, $f(-1)=1-1+left(frac{1}{2}-frac{1}{3}right)+left(frac{1}{4}-frac{1}{5}right)+frac{1}{6}>0$

The $f(x)>0$ and hence $f$ has no real zeros.
Now let, $g(x)=1+frac{x}{1}+frac{x^{2}}{2}+frac{x^{3}}{3}+cdots+frac{x^{7}}{7}$

An odd degree polynomial has at least one real root.
If our polynomial g has more than one zero, say $x_{1}, x_{2}$

Then by Role’s theorem in $left(x_{1}, x_{2}right)$ we have $^{prime} x_{3}$ ‘ such that $mathrm{g}^{prime}left(x_{3}right)=0$

$Rightarrow 1+x_{3}+x_{3}^{2}+cdots+x_{3}^{6}=0$

But this has no real zeros. Hence the given polynomial has exactly one real zero.

correct me if i Am wrong