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: 72

Answer

$\color{blue}{(3m+n+1)(3m-n-1)}$

Work Step by Step

Group the last three terms together to obtain: $9m^2-n^2-2n-1=9m^2-(n^2+2n+1)$ The trinomial is a perfect square trinomial in the form $a^2+2ab+b^2$ with $a=n$ and $b=1$. RECALL: $a^2-2ab+b^2=(a-b)^2$ Use the formula above to obtain: $=9m^2-(n+1)^2 \\=(3m)^2-(n+1)^2$ The polynomial above is a difference of two squares. Factor using the formula $a^2-b^2=(a+b)(a-b)$ with $a=3m$ and $b=n+1$ to obtain: $=[3m+(n+1)][3m-(n+1)] \\=\color{blue}{(3m+n+1)(3m-n-1)}$
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