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Unformatted text preview: sses through the x-axis at the points −
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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