Exercise 1111Algebra › PolynomialsL1Added 3 Nov 2026

Divisibility of polynomials

#divisibility#roots#factorisation

Statement

Show that $(X-1)^2$ divides $P(X)=X^n-nX+n-1$ for every $n\ge1$ .

Hint

Show $P(1)=0$ and $P'(1)=0$ .

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Value at $1$ . $P(1)=1-n+n-1=0$ .

Derivative. $P'(X)=nX^{n-1}-n$ , so $P'(1)=n-n=0$ .

Conclusion. $P(1)=P'(1)=0$ , so $1$ is a root of multiplicity at least $2$ , i.e. $(X-1)^2\mid P(X)$ .

Factor form. $P(X)=(X-1)^2 Q(X)$ with $Q$ of degree $n-2$ .