The equation $$x^{3} + (2r + 1)x^{2} + (4r - 1)x + 2 =0$$ has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is

Given that -2 is a root of the given cubic equation.

=> Dividing the given equation by (x + 2), Using the Horners method of synthetic division:

coefficient of $$x^2$$ is 1, and coefficient of x is (2r+1)-2 = 2r-1 and the constant term = (4r-1)-2(2r-1) = 1.

=> The quadratic obtained by dividing the cubic = $$x^2+\left(2r-1\right)x+1=0$$, Since, this equation has 2 real roots => Discriminant should be greater than 0

=> $$\left(2r-1\right)^2>4$$ => 2r-1 > 2 or 2r-1 < -2 => r > 3/2 or r < -1/2.

=> Minimum possible non-negative integer value of r is 2.