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Exercise 1825Algebra › PolynomialsOlympiadAdded 3 Nov 2026

Roots and coefficients

#roots#coefficients#bounds

Statement

Let P(X)=Xn+an−1Xn−1+⋯+a0P(X)=X^n+a_{n-1}X^{n-1}+\cdots+a_0 be monic with roots α1,…,αn\alpha_1,\ldots,\alpha_n (complex). Show that each ∣αi∣≤1+max⁡∣aj∣|\alpha_i|\le1+\max|a_j|.

Hint

If ∣α∣>1|\alpha|>1, then ∣α∣n≤∑j=0n−1∣aj∣∣α∣j|\alpha|^n\le\sum_{j=0}^{n-1}|a_j||\alpha|^j.

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Solution

Setup. If α\alpha is a root, P(α)=0P(\alpha)=0, so
αn=−(an−1αn−1+⋯+a0).\alpha^n=-(a_{n-1}\alpha^{n-1}+\cdots+a_0).

Bound. Let M=max⁡j∣aj∣M=\max_j|a_j|. If ∣α∣>1|\alpha|>1:
∣α∣n≤M(∣α∣n−1+⋯+1)=M∣α∣n−1∣α∣−1≤M∣α∣n∣α∣−1.|\alpha|^n\le M(|\alpha|^{n-1}+\cdots+1)=M\frac{|\alpha|^n-1}{|\alpha|-1}\le M\frac{|\alpha|^n}{|\alpha|-1}.

Conclude. Dividing by ∣α∣n|\alpha|^n: 1≤M/(∣α∣−1)1\le M/(|\alpha|-1), i.e. ∣α∣≤M+1|\alpha|\le M+1.

Conclusion. All roots satisfy ∣α∣≤1+M|\alpha|\le1+M with M=max⁡∣aj∣M=\max|a_j|.