Exercise 1827Algebra › PolynomialsL1Added 3 Nov 2026

A polynomial with parameter

#parameter#roots#Vieta

Statement

Find $m$ such that $P(X)=X^3-mX^2+X-1$ has a root at $X=1$ , then factor $P$ completely.

Hint

$P(1)=0$ gives $m=1$ . Then factor.

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$P(1)=0$ . $1-m+1-1=0$ , so $m=1$ .

Factor. $P(X)=X^3-X^2+X-1=X^2(X-1)+(X-1)=(X-1)(X^2+1)$ .

Conclusion. For $m=1$ , $P(X)=(X-1)(X^2+1)$ . Roots: $1$ , $i$ , $-i$ over $\mathbb{C}$ ; only $1$ over $\mathbb{R}$ .