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Answer:
(x -4)(x +1)
Step-by-step explanation:
[tex]\dfrac{x^2-16}{x^2+5x+4}=\dfrac{(x+4)(x-4)}{(x+4)(x+1)}=\boxed{\dfrac{x-4}{x+1}}\qquad x\ne-4[/tex]
The simplified form comes with a domain restriction corresponding to the hole in the function where the original denominator is zero.