Find Possible values of $c$ such that polynomial $$x(x+1)(x+2) \cdots (x+2009)=c$$ has one root of multiplicity two

My Try:

Assuming the polynomial equation has root $k$ Repeated twice The given polynomial can be written as

$$f(x)=x(x+1)(x+2)\cdots (x+2009)-c =(x+k)^2 \left(x^{2008}+a_1x^{2007}+a_2x^{2006}+\cdots-\frac{c}{k^2}\right)$$

Now $f'(x)=0$ should have $k$ as a root with multiplicity One

we have


can we use this concept to find $c$

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    $\begingroup$ note that the polynomial with $c=0$ has all different roots so the intersection with line $y=c$ will have only single roots unless $f'(x)=0$ i.e. it is a local extremum and $y=c$ is tangent in this abscissa. So you are good for searching for all these... $\endgroup$ – zwim Apr 28 '18 at 16:40
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    $\begingroup$ Is the intended question to ask "How many possible values?" I think it is not feasible to find an explicit form for $c$. $\endgroup$ – Jeffery Opoku-Mensah Apr 28 '18 at 16:45

A plot of $x(x+1)(x+2)(x+3)$ is below. Your polynomial will have lots more squiggles but be similar. You get a double root any time $c$ matches one of the local maxima or minima. You should be able to convince yourself that your polynomial is symmetric around $x=-1004.5$. If you want exactly one double root, that will only happen when $c$ matches the central peak. The value there is $$\prod_{n=0}^{1004}\left(n+\frac 12\right)^2 \approx 5.28\cdot 10^{5161}$$If you want at least one double root there will be $1004$ more values, which I think will be hard to find.

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