Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 8 - Rational Functions - Chapter Test - Page 557: 19

Answer

$\frac{x^3 + 6x^2 + 9x - 1}{(x + 2)(x - 2)}$ Restriction: $x \ne -2, 2$

Work Step by Step

Factor all expressions in the original exercise: $\dfrac{x(x + 4)}{x - 2} + \dfrac{x - 1}{(x + 2)(x - 2)}$ Before performing the addition, find the least common denominator (LCD) of the two fractions. The LCD incorporates all factors in both denominators. Then convert the original fractions to equivalent fractions using the LCD, which is $(x+2)(x-2)$: $\dfrac{x(x + 4)(x + 2)}{(x + 2)(x - 2)} + \dfrac{x - 1}{(x + 2)(x - 2)}$ Use distributive property and then multiply the binomials: $\dfrac{x(x^2 + 6x + 8)}{(x + 2)(x - 2)} + \dfrac{x - 1}{(x + 2)(x - 2)}$ Multiply to simplify: $\dfrac{x^3 + 6x^2 + 8x}{(x + 2)(x - 2)} + \dfrac{x - 1}{(x + 2)(x - 2)}$ Add the fractions: $\dfrac{x^3 + 6x^2 + 8x + x - 1}{(x + 2)(x - 2)}$ Combine like terms: $\dfrac{x^3 + 6x^2 + 9x - 1}{(x + 2)(x - 2)}$ Restrictions on $x$ occur when the value of $x$ makes the fraction undefined, which means that the denominator becomes $0$. Set the factors in the denominators equal to $0$ to find restrictions: First factor: $x + 2 = 0$ Subtract $2$ from each side of the equation: $x = -2$ Second factor: $x - 2 = 0$ Add $2$ to each side of the equation: $x = 2$ Restriction: $x \ne -2, 2$
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