Cubics

\(\text{you can see the zero crossings are at $x=0, x=1, x=2$}\\ \text{Thus the curve is given by }\\ f(x) = x(x-1)(x-2) = x^3-3 x^2+2 x\\ a+b+c+d =1-3+2= 0\)

\(\text{one can be a bit more clever and note that in general}\\ a+b+c+d = f(1)\\ \text{and we know that $1$ is one of the zeros so}\\ a+b+c+d=f(1)=0\)