College Algebra (6th Edition)

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

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 86: 25

Answer

$\frac{(x-2)^{2}}{x}; x \ne 0,2,-2$

Work Step by Step

$(x^{2}-4) = (x^{2}-2^{2})$ is the difference of squares so the factors are $(x+2)(x-2)$ The given expression is written as $\frac{(x+2)(x-2)}{x}\div \frac{(x+2)}{(x-2)} = \frac{(x+2)(x-2)}{x}\times \frac{(x-2)}{(x+2)} $ Exclude the values 0, 2 and -2 for $x.$ Otherwise they make the denominator zero. $= \frac{(x+2)(x-2)}{x}\times \frac{(x-2)}{(x+2)} ; x \ne 0,2,-2$ Divide numerator and denominator by common factors. $=\frac{(x-2)^{2}}{x}; x \ne 0,2,-2$
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