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: 62

Answer

$$\color{blue}{ f(x)= x^2-14x+53 }$$

Work Step by Step

We are given zeros and asked to write a polynomial function. If $ 7-2i $ is a zero, then $n +7-2i =0$ $n -2i = -7 $ $n= -7 +2i $ So our factor is $\bf{(x -7 +2i )}$ because when $x= 7-2i $, $(x -7 +2i )=0$ If $ 7+2i $ is a zero, then $n 7+2i =0$ $n +2i = -7 $ $n= -7-2i $ So our factor is $\bf{(x -7-2i )}$ because when $x= 7+2i $, $(x -7-2i )=0$ So our function is: $f(x)= (x -7 +2i )(x -7-2i ) $ $f(x)= (x^2-7x-2ix-7x+49+14i+2ix-14i-4i^2) $ $f(x)= (x^2-14x+49-4i^2) $ Recall that $i^2=-1$ $f(x)= (x^2-14x+49-4(-1)) $ $f(x)= (x^2-14x+49+4) $ $$\color{blue}{ f(x)= x^2-14x+53 }$$
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