Suppose the derivative of function f is f'(x) =(x+1)2(x-3)5(x-6)4 On what interval is f increasing?

Assuming f'(x)=(x+1)^2(x-3)^5(x-6)^4 (if you meant f'(x)=40(x+1)(x-3)(x-6) see below), we are asked to find the interval(s) when f is increasing.
Since f is a polynomial, it is continuous and differentiable everywhere, so f is increasing whenever f'(x)>0.
f'(x) has zeroes at x=-1,x=3, and x=6. Note that the zeros at x=-1 and x=6 have even multiplicity so the derivative does not change sign at these points. We test the derivative to the left and right of 3: f'(0)=-314928<0 and f'(4)=400 >0. Note that f'(6)=0.
The derivative is greater than or equal to zero for all x>3.
Thus the function is strictly monotonic (strictly increasing) on (3,oo).
The graph of the derivative:

The function decreases on (-oo,3) where the rate of decrease slows at x=-1; the function grows from 3 onward—the rate of growth slows to zero at x=6 and resumes.
If the multiplicity of a root of a polynomial function is even (the root is a double root, quadruple root, etc.), the function does not change sign at that root. If the multiplicity is odd, the function changes sign (i.e., it goes from positive to negative or vice versa).
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If you meant f'(x)=40(x+1)(x-3)(x-6), then f'(x)>0 on (-1,3) and (6,oo), and these are the intervals where the function would be increasing .
http://mathworld.wolfram.com/MonotonicFunction.html

http://mathworld.wolfram.com/MultipleRoot.html