Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Review Exercises: 36

Answer

$(x+1)(x-1)(x-2)(x^2+2x+4)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ x^5-x^3-8x^2+8 ,$ use factoring by grouping. This results to a factor that is a difference of $2$ squares and a factor that is a difference of $2$ cubes. Use the appropriate factoring method for each factor. $\bf{\text{Solution Details:}}$ Using factoring by grouping, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^5-x^3)-(8x^2-8) \\\\= x^3(x^2-1)-8(x^2-1) \\\\= (x^2-1)(x^3-8) .\end{array} The expressions $ x^2 $ and $ 1 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^2-1 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} [(x)^2-(1)^2](x^3-8) \\\\= (x+1)(x-1)(x^3-8) .\end{array} The expressions $ x^3 $ and $ 8 $ are both perfect cubes (the cube root is exact). Hence, $ x^3-8 $ 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 above is equivalent to \begin{array}{l}\require{cancel} (x+1)(x-1)[(x)^3-(2)^3] \\\\= (x+1)(x-1)[(x-2)(x^2+2x+4)] \\\\= (x+1)(x-1)(x-2)(x^2+2x+4) .\end{array}
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