Indian Statistical Institute B.Math & B.Stat : Roots of Polynomial

Let $c$ be a fixed real number. Show that a root of the equation $$x(x+1)(x+2)\dots(x+2009)=c$$ can have multiplicity at most $2.$ Let $f(x) = x(x+1)(x+2)\dots(x+2009)-c$ First we compute the derivative of $f(x)$ and see that $f'(x) = (x+1)(x+2)\dots(x+2009)+x(x+2)\dots(x+2009)+\dots+x(x+1)\dots(x+r-1)(x+r+1)\dots(x+2009)+$ $\dots+x(x+1)(x+2)\dots(x+2008)$ where $r$ is a positive integer less than $2009$. Now $f'(-r) = (-r)(-r+1)\dots(-1)(1)\dots(-r+2009) =(-1)^r r! (2009-r)! > 0$ $if$ $r$ is even, else $<0$.where $r \in \{0,1,2,\dots,2008\}$ Thus we have the following inequalities, $f'(0) > 0, f'(-1) < 0, f'(-2) > 0, \dots, f'(2008) > 0, f'(2009) < 0$ This shows that $f'(x)=0$ has one real root in each of the intervals $(-1,0),(-2,-1),\dots,(-2009,-2008)$. Since $degree$ of $f'(x)$ is $2009$, all the roots of $f'(x)=0$ is real and simple. Thus a root of $f'(x)=0$ cannot be a root of the equation $f''(x)=0$. So a root of $f(x)=0$ can have $multiplicity$ at most $2$.