Stitz-Zeager_College_Algebra_e-book

Since we have the same denominator in both terms we

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Unformatted text preview: rn to the business of zeros. Suppose we wish to find the zeros of f (x) = x2 − 2x + 5. To solve the equation x2 − 2x + 5 = 0, we note the quadratic doesn’t factor nicely, so we resort to the Quadratic Formula, Equation 2.5 and obtain √ −(−2) ± (−2)2 − 4(1)(5) 2 ± −16 2 ± 4i x= = = = 1 ± 2i. 2(1) 2 2 Two things are important to note. First, the zeros, 1 + 2i and 1 − 2i are complex conjugates. If ever we obtain non-real zeros to a quadratic function with real coefficients, the zeros will be a complex conjugate pair. (Do you see why?) Next, we note that in Example 3.4.1, part 6, we found (x − [1 + 2i])(x − [1 − 2i]) = x2 − 2x + 5. This demonstrates that the factor theorem holds even for non-real zeros, i.e, x = 1 + 2i is a zero of f , and, sure enough, (x − [1 + 2i]) is a factor of f (x). It turns out that polynomial division works the same way for all complex numbers, real and non-real alike, and so the Factor and Remainder Theorems hold as well. But how do we know if...
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