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If we wanted to completely factor f x over the real

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Unformatted text preview: sses through the x-axis at the points − √ 6 2 ,0 and 6 2 ,0 . Since −1 is a zero of multiplicity 2, we know the graph of y = f (x) touches and rebounds off the x-axis at (−1, 0). Putting this together, we get y x You can see why the ‘no calculator’ approach is not very popular these days. It requires more computation and more theorems than the alternative.4 In general, no matter how many theorems 4 This is apparently a bad thing. 3.3 Real Zeros of Polynomials 215 you throw at a polynomial, it may well be impossible5 to find their zeros exactly. The polynomial f (x) = x5 − x − 1 is one such beast.6 According to Descartes’ Rule of Signs, f has exactly one positive real zero, and it could have two negative real zeros, or none at all. The Rational Zeros Test gives us ±1 as rational zeros to try but neither of these work since f (1) = f (−1) = −1. If we try the substitution technique we used in Example 3.3.4, we find f (x) has three terms, but the exponent on the x5 isn’t e...
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