Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.5 Special Factoring Techniques - 3.5 Exercises - Page 279: 26



Work Step by Step

The student incorrectly used the factoring of the difference of $2$ squares. The factoring of the difference of $2$ cubes should instead be used. The expressions $ 8x^3 $ and $ 27 $ are both perfect cubes (the cube root is exact). Hence, $ 8x^3-27 $ is a $\text{ difference }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes, which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$, the expression, $8x^3-27,$ is equivalent to \begin{array}{l}\require{cancel} (2x)^3-(3)^3 \\\\= (2x-3)[(2x)^2+2x(3)+(3)^2] \\\\= (2x-3)(4x^2+6x+9) .\end{array}
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