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Stitz-Zeager_College_Algebra_e-book

# 1 x 3 mult 1 x 5 mult 1 x 2 3 3 5 mult 1 n

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Unformatted text preview: chy’s Bound, we ﬁnd M = 12, so all of the real zeros lie in the interval [−13, 13]. 2. Applying the Rational Zeros Theorem with constant term a0 = −12 and leading coeﬃcient a4 = 1, we get the list {± 1, ± 2, ± 3, ± 4, ± 6, ± 12}. 3. Graphing y = f (x) on the interval [−13, 13] produces the graph below on the left. Zooming in a bit gives the graph below on the right. Based on the graph, none of our rational zeros will work. (Do you see why not?) 3.3 Real Zeros of Polynomials 211 4. From the graph, we know f has two real zeros, one positive, and one negative. Our only hope at this point is to try and ﬁnd the zeros of f by setting f (x) = 0 and solving. Doing so results in the equation x4 + x2 − 12 = 0. If we stare at this equation long enough, we may recognize it as a ‘quadratic in disguise’.2 In other words, we have three terms: x4 , x2 , and 12, and the exponent on the ﬁrst term, x4 , is exactly twice that of the second term, x2 . We may rewrite 2 this as x2 + x2 − 12 = 0. To better see the forest fo...
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