SECTION 4.4 Complex Zeros; Fundamental Theorem of Algebra 233 We can now prove the theorem stated in Section 4.3. THEOREM Every polynomial function with real coefficients can be uniquely factored over the real numbers into a product of linear factors and/or irreducible quadratic factors. Proof Every complex polynomial function f of degree n has exactly n zeros and can be factored into a product of n linear factors. If its coefficients are real, the zeros that are complex numbers always occur in conjugate pairs. As a result, if = + r a bi is a complex zero, then so is = − r a bi. Consequently, when the linear factors x r − and x r − of f are multiplied, we have ( )( ) ( ) − −=−+ +=− ++ x r x r x r r x rr x ax a b 2 2 2 2 2 This second-degree polynomial has real coefficients and is irreducible (over the real numbers). So, the factors of f are either linear or irreducible quadratic factors. ■ Finding the Complex Zeros of a Polynomial Function Find the complex zeros of the polynomial function ( ) = + + + − f x x x x x 3 5 25 45 18 4 3 2 Write f in factored form. Solution EXAMPLE 3 Step 1 The degree of f is 4. So f has four complex zeros. From Descartes’ Rule of Signs, there is one positive real zero. Also, since ( ) − = − + − − f x x x x x 3 5 25 45 18, 4 3 2 there are three negative real zeros or one negative real zero. Step 2 The Rational Zeros Theorem provides information about the potential rational zeros of polynomial functions with integer coefficients. For this polynomial function (which has integer coefficients), the potential rational zeros are ± ± ± ± ± ± ± ± 1 3 , 2 3 , 1, 2, 3, 6, 9, 18 Step 3 Figure 34 shows the graph of f using a TI-84 Plus CE. The graph has the characteristics expected of this polynomial function of degree 4: It behaves like = y x3 4 for large x and has y -intercept −18. There are x -intercepts near −2 and between 0 and 1. Step 4 Because ( ) − = f 2 0, we know that −2 is a zero of f and ( ) − − = + x x 2 2 is a factor of f . Use long division or synthetic division to factor f . Using synthetic division, ) − − − − − − 2 3 5 25 45 18 6 2 54 18 3 1 27 9 0 So ( ) ( ) ( ) = + − + − f x x x x x 2 3 27 9 3 2 . The depressed equation is ( ) = − + − = q x x x x 3 27 9 0 1 3 2 Figure 34 ( ) = + + + − f x x x x x 3 5 25 45 18 4 3 2 45 245 23 1 Now Work PROBLEM 19 3 Find the Complex Zeros of a Polynomial Function The steps for finding the complex zeros of a polynomial function are the same as those for finding the real zeros. (continued)
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