## College Algebra (10th Edition)

Published by Pearson

# Chapter 5 - Section 5.6 - Complex Zeros; Fundamental Theorem of Algebra - 5.6 Assess Your Understanding: 19

#### Answer

$\color{blue}{P(x)=x^5-4x^4+7x^3-8x^2+6x-4}$

#### Work Step by Step

RECALL: If $a+bi$ is a zero of a polynomial function with real number coefficients, then its conjugate $a-bi$ is also a zero of the function. Thus, the missing zeros of the given function are the conjugates of $-i$ and $1+i$, which are ${\bf i, 1-i}$. The zeros of the polynomial function with real coefficients are: $2 \\-i \\i \\1+i \\1-i$ This means that the function is: $P(x) = a(x-2)(x-i)(x+i) \cdot [x-(1+i)][x-(1-i)] \\P(x) = a(x-2)(x^2-i^2)(x-1-i)(x-1+i) \\P(x) =a(x-2)(x^2-(-1)] \cdot [(x-1)-i][(x-1)+i)]] \\P(x) = a(x-2)(x^2+1)[(x-1)^2-i^2] \\P(x) = a(x-2)(x^2+1)[x^2-2x+1-(-1)] \\P(x) = a(x-2)(x^2+1)(x^2-2x+1+1) \\P(x)=a(x-2)(x^2+1)(x^2-2x+2) \\P(x) = a(x-2)(x^4-2x^3+2x^2+x^2-2x+2) \\P(x)=a(x-2)(x^4-2x^3+3x^2-2x+2) \\P(x)=a(x^5-2x^4+3x^3-2x^2+2x-2x^4+4x^3-6x^2+4x-4) \\P(x)=a(x^5-4x^4+7x^3-8x^2+6x-4)$ Setting the leading coefficient $a=1$ gives: $\color{blue}{P(x)=x^5-4x^4+7x^3-8x^2+6x-4}$

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