Exercise · ZW-FUNDAMENTAL-THEOREM-OF-ALGEBRA-005 · ★★☆ Intermediate

The roots of \(z^3-6z^2+11z-6\) add up to:

Choose one option.

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  • C. \(6\)

Explanation

With \(p=\prod(z-z_k)\), the coefficient of \(z^2\) is \(-\sum z_k\). Here it is \(-6\), so the sum is \(6\) (roots \(1,2,3\)).

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Review the concept: Fundamental theorem of algebra

Every non-constant polynomial \(p\) with complex coefficients, of degree \(n\), has at least one root in \(\mathbb{C}\); hence it splits into \(n\) linear factors \(z-z_k\), with leading coefficient \(a_n\).

\[p(z)=a_n\prod_{k=1}^{n}(z-z_k)\]

More practice problems: fundamental theorem of algebra

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