Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 3 - Polynomial and Rational Functions - 3.3 Zeros of Polynomial Functions - 3.3 Exercises - Page 337: 69

Answer

$$\color{blue}{f(x)= x^3-8x^2+22x-20 }$$

Work Step by Step

We are given zeros and asked to write a polynomial function. If $ 2 $ is a zero, then $n +2 =0$ $n= -2 $ So our factor is $\bf{(x-2 )}$ because when $x= 2 $, $(x -2 )=0$ If $ 3+i $ is a zero, then $n +3+i =0$ $n +i = -3 $ $n= -3-i $ So our factor is $\bf{(x -3-i )}$ because when $x= 3+i $, $(x-3-i )=0$ Since $ 3+i $ is a complex number, its conjugate, $ 3-i $ is also a zero $n +3-i =0$ $n -i =-3$ $n= -3+i $ So our factor is $\bf{(x -3+i )}$ because when $x= 3-i $, $(x -3+i )=0$ So our function is: $f(x)=(x-2)(x -3-i)(x -3+i)$ $f(x)=(x-2)(x^2-6x+9-i^2)$ Recall that $i^2=-1$ $f(x)=(x-2)(x^2-6x+9-(-1))$ $f(x)=(x-2)(x^2-6x+10)$ $$\color{blue}{f(x)= x^3-8x^2+22x-20 }$$
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