Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Review - Test - Page 323: 8

Answer

$(x-1)^2(x-2i)(x+2i)$

Work Step by Step

Try to factor P as $P(x)=x^4-x^3-x^3+x^2+4x^2-4x-4x+4=x^3(x-1)-x^2(x-1)+4x(x-1)-4(x-1)=(x-1)(x^3-x^2+4x-4)=(x-1)[x^2(x-1)+4(x-1)]=(x-1)^2(x^2+4)=(x-1)^2(x-2i)(x+2i)$
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