College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.4 - Factoring Polynomials - R.4 Exercises: 64


(2m - 3n)(4$m^{2}$ + 6mn + 9$n^{2}$)

Work Step by Step

8$m^{3}$ - 27$n^{3}$ is a difference of cubes, so we write it as a difference of cubes and factor: $(2m)^{3}$ - $(3n)^{3}$ = (2m - 3n)($(2m)^{2}$ + 2m(3n) + $(3n)^{2}$) = (2m - 3n)(4$m^{2}$ + 6mn + 9$n^{2}$)
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