College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Summary, Review, and Test - Review Exercises: 107

Answer

= $(2 + x)(2 - x)$$(x^{2} + 3)^{\frac{1}{2}}$ [$x^{4} - x^{2} - 13 $]

Work Step by Step

$(x^{2} - 4)(x^{2} + 3)^{\frac{1}{2}}$ - $(x^{2} - 4)^{2}(x^{2} + 3)^{\frac{3}{2}}$ = $(x^{2} - 4)(x^{2} + 3)^{\frac{1}{2}}$ - $(x^{2} - 4)(x^{2} - 4)(x^{2} + 3)^{\frac{1}{2} +1}$ = $(x^{2} - 4)(x^{2} + 3)^{\frac{1}{2}}$[1 - $(x^{2} - 4)(x^{2} + 3)$] = $(x^{2} - 4)(x^{2} + 3)^{\frac{1}{2}}$ [1 - ($x^{4} - 4x^{2} + 3x^{2} - 12$] = $(x^{2} - 4)(x^{2} + 3)^{\frac{1}{2}}$ [1 - $x^{4} + 4x^{2} - 3x^{2} + 12$] = $(x^{2} - 4)(x^{2} + 3)^{\frac{1}{2}}$ [13- $x^{4} + x^{2} $] = $(4 - x^{2})(x^{2} + 3)^{\frac{1}{2}}$ [$x^{4} - x^{2} - 13 $] by $(4 - x^{2})$ = $(2 + x)(2 - x)$ $(4 - x^{2})(x^{2} + 3)^{\frac{1}{2}}$ [$x^{4} - x^{2} - 13 $] = $(2 + x)(2 - x)$$(x^{2} + 3)^{\frac{1}{2}}$ [$x^{4} - x^{2} - 13 $]
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