Algebra 2 Common Core

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

Chapter 6 - Radical Functions and Rational Exponents - Chapter Review - Page 425: 56

Answer

$\frac{g(x)}{f(x)} = x + 4$ $\text{The domain is all real numbers except }4$.

Work Step by Step

In this problem, we are asked to divide one function from another. Let us go ahead and set up the division problem: $\dfrac{g(x)}{f(x)} = \dfrac{x^2 - 16}{x - 4}$ Let's factor the numerator to see if we can cancel one of the factors with the denominator. We can see that the binomial in the numerator is the difference of two squares. The factorization for the difference of two squares is as follows: $(a^2 - b^2) = (a - b)(a + b)$ Let us plug in $\sqrt {x^2}$ or $x$ for $a$ and $\sqrt {16}$ or $4$ for $b$: $\dfrac{g(x)}{f(x)} = \dfrac{(x - 4)(x + 4)}{x - 4}$ We can cancel out an $x - 4$ from both the numerator and denominator: $\dfrac{g(x)}{f(x)} = x + 4$ We cannot have $x = 4$ because if $x$ were $4$, the denominator would be $0$. The denominator cannot equal $0$ or else the function would be undefined. The domain is all real numbers except $4$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.