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 43: 24



Work Step by Step

$(z+4)$ is in both terms, we can factor it out: $(z+4)((3z+2)-(z+6))=(z+4)(2z-4)$ We can factor out a constant of 2 from the second term: $(z+4)(2z-4)=(z+4)\times2\times (z-2)=2(z+4)(z-2)$
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