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: 81

Answer

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

Work Step by Step

With $27=3^3$, the given polynomial is equivalent to: $=3^3-(m+2n)^3$ Factor using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=3$ and $b=m+2n$ to obtain: $=[3-(m+2n)][3^2+3(m+2n) + (m+2n)^2] \\=(3-m-2n)[9+3m+6n+(m^2+4mn+4n^2)] \\=\color{blue}{(3-m-2n)(9+3m+6n+m^2+4mn+4n^2)}$
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