Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises - Page 45: 130

Answer

$(x^4-x^2+1)(x^2-x+1)(x^2+x+1)$

Work Step by Step

Step 1. Let $y=x^2$, we have $x^8+x^4+1=y^4+y^2+1$ Step 2. Use the formula obtained from the previous exercise, we have $y^4+y^2+1=(y^2-y+1)(y^2+y+1)$ Step 3. Replace $y$ with $x^2$, we have $x^8+x^4+1=(x^4-x^2+1)(x^4+x^2+1)$ Step 4. Use the formula obtained from the previous exercise for the second term, we have $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$ Step 5. Thus $x^8+x^4+1=(x^4-x^2+1)(x^2-x+1)(x^2+x+1)$
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