By substituting values for x we can see the behaviour of f(x), in particular where f(x) goes from positive to negative or vice versa. That means there is a root between certain values of x. Between -1 and 0 such a root exists because f(-1)=-7 and f(0)=7. So x=-1 and 0 are the bounds of one root. Narrowing the bounds tells us that f(-0.9)=-2.288 and f(-0.8)=1.207, so there is a root between -0 9 and -0.8.
We can also see that f(2)=11, f(3)=1 and f(4)=63, indicating from this dip that there may be a real root between x=2 and 4. However, f(x) dips but doesn't change sign, so there are no real roots here.
If we use synthetic division assuming a near root of x=-1, we get (x^4-7x^3+16x^2-16x+14) with a remainder of -7, which clearly changes 14 into 7, the constant in f(x). This time x between 3 and 4 again appears to be in the vicinity of a root but there is no change of sign and closer inspection shows that the function does not intercept the axis.
Therefore the only integer bounds appear to be -1 to 0 or, if we go to one decimal place, -0.9 to -0.8.