Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 286: 110

Answer

The equation has one rational root: $~~x = -1$ The equation has either two or four irrational roots.

Work Step by Step

$P(x) = x^5-x^4-x^3-5x^2-12x-6 = 0$ We can list the possible rational zeros: $\frac{factors~of~6}{factors~of~1} = \pm 1, \pm 2, \pm 3, \pm 6$ Using synthetic division to check each possible rational zero, we find that only $x = -1$ is a rational zero. We can consider Descartes Rule of Signs. $P(x)$ has one variation in sign, so it has one positive real zero. $P(-x) = -x^5-x^4+x^3-5x^2+12x-6 = 0$ $P(-x)$ has four variations in sign, so $P(x)$ has four, two, or zero negative real zeros. We found above that $x = -1$ is a real zero, so $P(x)$ has four or two negative real zeros. Thus. $P(x)$ has one positive real zero and two or four negative real zeros. The only rational zero is $x = -1$, which is a negative real zero. Therefore, $P(x)$ has one positive irrational zero and one or three negative irrational zeros. $P(x)$ has either two or four irrational roots.
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