Which of the following is equal to the rational expression below when $x$ not equal to 4?
$\frac{x^{2} - 16}{x - 4}$

$\frac{x + 4}{x - 4}$

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$x - 4$

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$\frac{1}{x - 4}$

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$x + 4$

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Solution

The correct option is D
$x + 4$

Explanation for the correct option:
Given : $\frac{x^{2} - 16}{x - 4} [x \neq 4]$
Step-by-step explanation:
Factor the difference of squares on the numerator
$x^{2} - 16$ $= x^{2} - 4^{2} = (x - 4) (x + 4),$ thus expression becomes
$\begin{array}{rcl} \frac{x^{2} - 16}{x - 4} & = & \frac{(x - 4) (x + 4)}{x - 4} \\ = & x + 4 \end{array}$
Hence, the correct answer is option $(D)$ .

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