Question 245900
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*[tex \LARGE x\ +\ 4] is not a factor of *[tex \LARGE x^2\ +\ 8x\ -\ 16], so 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2\ +\ 8x\ -\ 16}{x\ +\ 4}]


is already reduced to lowest terms.


If you are asking for a quotient and remainder, you should have said so.  Write back and I'll tell you where to go look to find out how to do that.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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