Why draw graph when you can solve it directly? f(0) value is there already, f(4) can be calculated as we have f(x). ( f is of degree 3 all roots are there so we can get f(x) )

The graph of a degree 3 polynomial f(x) = a0 + a1x + a2(x^2) + a3(x^3), where a3 ≠ 0 is a cubic curve, as can be seen here https://en.wikipedia.org/wiki/... Now as given, the polynomial is zero at x = 1, x = 2 and x = 3, i.e. these are the only 3 real roots of this polynomial. Hence we can write the polynomial as f(x) = K (x-1)(x-2)(x-3) where K is some constant coefficient. Now f(0) = -6K and f(4) = 6K ( by putting x = 0 and x = 4 in the above polynomial ) and f(0)*f(4) = -36(k^2), which is always negative. Hence option A. We can also get the answer by just looking at the graph. At x < 1, the cubic graph (or say f(x) ) is at one side of x-axis, and at x > 3 it should be at other side of x-axis. Hence +ve and -ve values, whose multiplication gives negative.