## Precalculus (10th Edition)

Published by Pearson

# Chapter 4 - Polynomial and Rational Functions - 4.6 Complex Zeros; Fundamental Theorem of Algebra - 4.6 Assess Your Understanding - Page 240: 19

#### Answer

$f(x)=x^5-4x^4+7x^3-8x^2+6x-4$

#### Work Step by Step

If $a$ is a zero of a function with multiplicity $b$ then $(x-a)^b$ is a “multiplier” of the function. The Conjugate Pairs Theorem says that if a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. That is, if $a + bi$ is a zero then so is $a – bi$ and vice-versa. Therefore the zeros of the function are: $2$, $1\pm i$ and $\pm i$, hence the function could e.g. be: $a(x-(1+i))(x-(1-i))(x-i)(x-(-i))(x-2)\\=a(x-1-i)(x-1+i)(x-i)(x+i)(x-2)\\=a(x^5-4x^4+7x^3-8x^2+6x-4).$ If $a=1$, the function is: $f(x)=x^5-4x^4+7x^3-8x^2+6x-4$ The degree is $5$ which is good.

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