Answer:
(x+5) (x=3)
(X+5) (x+1)
Explanation:
A removeable discontinuity is always found in the denominator of a rational function and is one that can be reduced away with an identical term in the numerator. It is still, however, a problem because it causes the denominator to equal 0 if filled in with the necessary value of x. In my function above, the terms (x + 5) in the numerator and denominator can cancel each other out, leaving a hole in your graph at -5 since x doesn't exist at -5, but the x + 1 doesn't have anything to cancel out with, so this will present as a vertical asymptote in your graph at x = -1, a nonremoveable discontinuity.