Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 11 - Rational Expressions and Functions - Mid-Chapter Quiz - Page 678: 21


$p = (x+4)(x+5)$ = $x^2+9x + 20$

Work Step by Step

To find p, we must consider factor the denominator into $(x+4)(x-3)$. For the expression to simplify into $\frac{x+5}{x-3}$, then $(x+4)$ must divide from the numerator and the denominator. We already know $x+5$ is part of the numerator, and now we know that $(x+4)$ is also. So, $p = (x+4)(x+5)$ = $x^2+9x + 20$ $\frac{x^2+9x+20}{x^2+x-12} = \frac{(x+4)(x+5)}{(x+4)(x-3)} = \frac{x+5}{x-3}$
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