Roughly speaking this means that near x 1 the graph

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Unformatted text preview: our quotient polynomial qn has degree 0 so it’s a constant. This argument gives us the following factorization theorem. Theorem 3.14. Complex Factorization Theorem: Suppose f is a polynomial function with complex number coefficients. If the degree of f is n and n ≥ 1, then f has exactly n complex zeros, counting multiplicity. If z1 , z2 , . . . , zk are the distinct zeros of f , with multiplicities m1 , m2 , . . . , mk , respectively, then f (x) = a (x − z1 )m1 (x − z2 )m2 · · · (x − zk )mk . Note that the value a in Theorem 3.14 is the leading coefficient of f (x) (Can you see why?) and as such, we see that a polynomial is completely determined by its zeros, their multiplicities, and its leading coefficient. We put this theorem to good use in the next example. Example 3.4.2. Let f (x) = 12x5 − 20x4 + 19x3 − 6x2 − 2x + 1. 1. Find all complex zeros of f and state their multiplicities. 2. Factor f (x) using Theorem 3.14 Solution. 1. Since f is a fifth degree polynomial, we know we need to perform at least three successful divisions to get the quotient down to a quadratic function. At that point, we...
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