## College Algebra (11th Edition)

Published by Pearson

# Chapter R - Section R.4 - Factoring Polynomials - R.4 Exercises: 69

#### Answer

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

#### Work Step by Step

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

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