Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 294: 56

Answer

$-1,3,i,-i$

Work Step by Step

Consider $P(x)=x^4=2x^3-2x^2-2x-3=0$ In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows: $x^4=2x^3-2x^2-2x-3=(x+1)(x^3-3x^2+x-3)=0$ This can be re-written as:$(x+1)[x(x^2+1)-3(x^2+1)]=0$ This gives: $(x+1)(x^2+1)(x-3)=0$ Here, $x+1=0 \implies x=-1\\x-3=0 \implies x=3\\x^2+1=0 \implies x^2=-1 \\ or, x=\pm i$ Thus the zeroes of $P(x)$ are: $-1,3,i,-i$
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