Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Review Exercises - Page 143: 107

Answer

$-(x+2)(x-2)(x^{2}+3)^{1/2}(x^{4}-x^{2}-13)$

Work Step by Step

We can factor out $(x^{2}-4)$, which is a difference of squares, and equals $(x+2)(x-2)$ Manipulating exponents, we can write $ A^{3/2}=A^{1+1/2}=A\cdot A^{1/2}$, meaning that we can factor out $(x^{2}+3)^{1/2}$ as well Problem =$ (x+2)(x-2)(x^{2}+3)^{1/2}[1-(x^{2}-4)(x^{2}+3)]$ =$ (x+2)(x-2)(x^{2}+3)^{1/2}[1-x^{4}-3x^{2}+4x^{2}+12]$ =$ (x+2)(x-2)(x^{2}+3)^{1/2}[-x^{4}+x^{2}+13]$ Factor out -1 from the brackets, as the trinomial they hold is prime (we can't find two factors of 13 whose sum is 1). = $-(x+2)(x-2)(x^{2}+3)^{1/2}(x^{4}-x^{2}-13)$
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