f(x)=x^3+8x^2-4x-32
note that f(2)=0 so x=2 is one of the zero
dividing f(x) by (x-2)
the quotient is x^2+10x+16 which factors into (x+2)(x+8)
combining f(x)=(x-2)(x+2)(x+8)
the zeros are -8, -2 and 2
for f(x) to "turn around", it must be a max or min
f'(x)=3x^2+16x-4=0
at greatest zero, f'(2)<>0 so it crosses the x-axis
at second greatest zero, f'(-2)<>0 so it crosses the x-axis
at smallest zero, f'(-8)<>0 so it crosses the x-axis
