College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 3, Polynomial and Rational Functions - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises: 10

Answer

(a) $\left\{0, \dfrac{-1+\sqrt3 i}{2}, \dfrac{-1-\sqrt3 i}{2} \right\}$. (b) $P(x) = x(x^2+x+1)$

Work Step by Step

$\bf{(a) \text{ Zeros}}$ Factor the polynomial completely to obtain: $P(x) = x(x^2+x+1)$ Equate each factor to zero then solve each equation to obtain: \begin{array}{ccccc} &x=0 &\text{or} &x^2+x+1=0\end{array} The second equation can be solved using the quadratic formula $x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. The trinomial has $a=1, b=1, c=1$. Substitute these values into the quadratic formula above to obtain: $x=\dfrac{-1\pm \sqrt{1^2-4(1)(1)}}{2(1)} \\x=\dfrac{-1\pm \sqrt{1-4}}{2} \\x=\dfrac{-1\pm \sqrt{-3}}{2} \\x=\dfrac{-1\pm \sqrt{(-1)(3)}}{2} \\x=\dfrac{-1\pm i\cdot \sqrt3}{2} \\x=\dfrac{-1\pm \sqrt3 i}{2}$ Thus, the zeros of the function are: $\left\{0, \dfrac{-1+\sqrt3 i}{2}, \dfrac{-1-\sqrt3 i}{2} \right\}$. $\bf{(b) \text{ Completely Factored Form}}$ From part (a) above, the completely factored form of $P(x)$ is: $P(x) = x(x^2+x+1)$
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