How to show that a root of the equationx(x+1)(x+2)…….(x+2009)=c

Let denote byP(x)=x(x+1)(x+2)…….(x+2009)−cWe haveP(0)=P(−1)=⋯P(−2009)=−cand by Rolle's theorem there's0&gt;t0≻1&gt;⋯&gt;t2008≻2009s.t.P′(t0)=P′(t1)=⋯=P′(t2008)=0Since the polynomial P has the degree 2010 then the degree of P' is 2009 and then the ti are the roots of P'If c=0 then the roots of P are different from the roots of P' and there is not a multiple root.If c≠0, we apply the Rolle's theoem on P' and we find the roots si of P″ s.t.t0&gt;s0&gt;t1&gt;⋯&gt;s2007&gt;t2008and if we have P(α)=P′(α)=0 then α=ti0 and then α≠si∀i=0,…,2007 and hence P‴(α)≠0

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Jefferson Pacheco

Answered question

2022-04-06

How to show that a root of the equation

x (x + 1) (x + 2) \dots \dots . (x + 2009) = c

Answer & Explanation

Kaphefsceasiaf8w9

Beginner2022-04-07Added 9 answers

Let denote by

P (x) = x (x + 1) (x + 2) \dots \dots . (x + 2009) - c

We have

P (0) = P (- 1) = \dots P (- 2009) = - c

and by Rolle's theorem there's

0 > t_{0} ≻ 1 > \dots > t_{2008} ≻ 2009

s.t.

P^{'} (t_{0}) = P^{'} (t_{1}) = \dots = P^{'} (t_{2008}) = 0

Since the polynomial P has the degree 2010 then the degree of P' is 2009 and then the

t_{i}

are the roots of P'
If

c = 0

then the roots of P are different from the roots of P' and there is not a multiple root.
If

c \neq 0

, we apply the Rolle's theoem on P' and we find the roots

s_{i}

P^{″}

s.t.

t_{0} > s_{0} > t_{1} > \dots > s_{2007} > t_{2008}

and if we have

P (α) = P^{'} (α) = 0

then

α = t_{i_{0}}

and then

α \neq s_{i} \forall i = 0, \dots, 2007

and hence

P^{‴} (α) \neq 0

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