Lesson: _____ Section 2.5 The Fundamental Theorem of Algebra

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Presentation transcript:

Lesson: _____ Section 2.5 The Fundamental Theorem of Algebra Counting complex & repeated solutions, an nth degree polynomial function has exactly n zeros. (*double roots count twice!) If a + bi is a zero, then a-bi is also a zero.  Complex zeros always travel in conjugate pairs.

Express as a product of linear factors and find all zeros. f(x) = x5 + x3 + 2x2 – 12x + 8

Ex. Given that (-2 + i) is a zero of f(x), find the remaining zeros. f(x) = x4 + 4x3 + 12x2 + 28x + 35

Distributing this would produce x2 – 3x – xi + 3i No need to write this down. Just follow along… Why do the complex roots have to travel in pairs? What if x = 3 and x = i were the only roots? Then (x – 3) and (x – i) would be factors. (x – 3)(x – i) Distributing this would produce x2 – 3x – xi + 3i …which is not a polynomial! (all polynomials must have real coefficients) The only way for an imaginary root to be acceptable is if somehow the i’s disappeared. This only occurs when you multiply conjugates.