Use synthetic division to get the cubic: x^3-x^2-6x+18
-2 | 1 1 -8 6 36
......1 -2 2 12 -36
......1 -1 -6 18 | 0
If the cubic factorises, the factors must have a product of 18: 1, 2, 3, 6, 9, 18 are the only factors. Try the smallest first, 1 or -1: 1-1-6+18 is not zero, so 1 is not a zero: -1-1+6+18 non-zero so -1 not a zero. Now 2 and -2: 8-4-12+18 non-zero; -8-4+12+18 non-zero; now 3 and -3: 27-9-18+18 non-zero; -27-9+18+18=0 so -3 is a zero and x+3 is a factor.
-3 | 1 -1 -6 18
......1 -3 12 -18
......1 -4 6 | 0
x^2-4x+6 is a quadratic which factorises no further into real factors, so the factors are (x+2)(x+3)(x^2-4x+6).
The curve cuts the x axis at -3 and -2 only. When x is large and negative f(x) is large and positive so the curve comes from the second quadrant cutting the x axis at -3 and -2 between which there will be a minimum. When x=0 f(x)=36 so the curve cuts the vertical axis at this point. There are two more turning points, a maximum followed by a minimum before the curve continues in the first quadrant, where x and f(x) become both large and positive. The sum of x^4 and x^3 mean that the graph climbs quite rapidly when x is positive.
To find the turning points means solving f'(x)=0=4x^3+3x^2-16x+6. There must be a zero between -2 and -3. So if we put x=-2.5 we should be fairly close: -4*15.625+3*6.25+16*2.5+6=-62.5+18.75+40+6=64.75-62.5=2.25. f(-2.5)=39.0625-15.625-50-15+6=45.0625-80.625 which is approx -35.5. This gives some idea how deep and steep the minimum is. We know the curve cuts the vertical axis at 36, when x=0. f(1)=44-8=36. This is the same number as the intercept, so there must be a little wriggle between x=0 and 1, accounting for the maximum and minimum.