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 3.2 - Polynomial Functions and Their Graphs - 3.2 Exercises - Page 302: 42

Answer

$P(x)=(x-2)^2(x^2+2x+4)$ Zero: $2 \text{ (multiplicity 2)}$ Refer to the graph below.
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Work Step by Step

Factor the polynomial completely to obtain: \begin{align*} P(x)&=(x^4-2x^3)+(-8x+16)\\ &=x^3(x-2)+(-8)(x-2)\\ &=(x-2)(x^3-8)\\ &=(x-2)(x-2)(x^2+2x+4)\\ &=(x-2)^2(x^2+2x+4)\\ \end{align*} To find the zeros, use the Zero-Product Property by equating each factor to $0$, then solve each equation to obtain: \begin{align*} (x-2)^2&=0 &\text{or}& &x^2+2x+4=0\\ x-2&=0 &\text{or}& &\text{(no real solution)} \\ x&=2 \text{ (multiplicity 2)}\\ \end{align*} The zero of the function is $2$. Use a graphing utility to graph $P(x)$. Refer to the graph above.
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