Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.1 - The Greatest Common Factor and Factoring by Grouping - Exercise Set - Page 428: 110


Makes sense.

Work Step by Step

Try an example: $2x^{3}-4x^{2}+7x-14$ $\left[\begin{array}{lll} (2x^{3}-4x^{2})+(7x-14) & ... & (2x^{3}+7x)+[-4x^{2}-14] \\ =2x^{2}(x-2)+7(x-2) & & =x(2x^{2}+7)-2(2x^{2}+7)\\ =(x-2)(2x^{2}+7) & & =(2x^{2}+7)(x-2) \end{array}\right]$ which are equivalent, since multiplication is commutative. Two different groupings, same factorizations. Makes sense.
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