Question

Pls help on this one?

Answer 1

Answer:

$\large\boxed{D.\ x+4}$

Step-by-step explanation:

$\dfrac{x^2-16}{x-4}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2}{x-4}=\dfrac{(x-4)(x+4)}{x-4}\\\\\text{canceled}\ (x-4)\\\\=\dfrac{x+4}{1}=x+4$