College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 5 - Section 5.6 - Complex Zeros; Fundamental Theorem of Algebra - 5.6 Assess Your Understanding - Page 394: 18



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 $1+2i$, which are ${\bf-1, 1-2i}$. The zeros of the polynomial function with real coefficients are: $i \\1+2i \\-i \\1-2i$ This means that the function is: $P(x) = a(x-i)(x+i) \cdot [x-(1+2i)][x-(1+2i)] \\P(x) = a(x^2-i^2)(x-1-2i)(x-1+2i) \\P(x) =a(x^2-(-1)] \cdot [(x-1)-2i][(x-1)+2i)]] \\P(x) = a(x^2+1)[(x-1)^2-4i^2] \\P(x) = a(x^2+1)[x^2-2x+1-4(-1)] \\P(x) = a(x^2+1)(x^2-2x+1+4) \\P(x)=a(x^2+1)(x^2-2x+5) \\P(x) = a(x^4-2x^3+5x^2+x^2-2x+5) \\P(x)=a(x^4-2x^3+6x^2-2x+5))$ Setting the leading coefficient $a=1$ gives: $\color{blue}{P(x)=x^4-2x^3+6x^2-2x+5}$
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