pls help on this one?
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Answer:
[tex]\large\boxed{D.\ x+4}[/tex]
Step-by-step explanation:
[tex]\dfrac{x^2-16}{x-4}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2}{x-4}=\dfrac{(x-4)(x+4)}{x-4}\\\\\text{canceled}\ (x-4)\\\\=\dfrac{x+4}{1}=x+4[/tex]