Question :

If :

$P(x)=x^3+ax^2+bx+c$ where $(a,b,c)\in Z^3$

And $m,n,k$ root of $P(x)$

such that : $m.n=k$

Then show that : $2P(-1)$ multiple of


My try :

We known that :




Since $m.n=k$

So : $k^2=-c$

But how I complete this work

Please give me ideas or hints


First let's simplify the things that are required in the proof.

$$P(1) + P(-1) - 2[1+P(0)] = (1+a+b+c)+(-1+a-b+c) -2(1+c)$$



$$2P(-1) = 2(-1+a-b+c)$$

Now we have to show that $2a-2$ divides $2(-1+a-b+c)$ which is equivalent to saying that $a-1$ divides $b-c$.(Why?)

Now writing out the expressions for $a,b,c$.

$$a = -(m+n+k)$$ $$b=mn+nk+mk$$ $$c = -mnk$$


$$b-c= mn+nk+mk+mnk$$ $$=k+nk+mk+k^2$$


Which is what we wanted!

  • $\begingroup$ Very good ideas , brilliant Sir $\endgroup$ – Kînan Jœd May 18 at 0:13
  • $\begingroup$ Sure, no problem. Plz mark the answer as accepted if you have no further queries. $\endgroup$ – Vizag May 18 at 17:46

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