Cubic Polynomial Real Roots Condition

Since $x = -2$ is a root, factor out $(x + 2)$: $x^3 + (2r+1)x^2 + (4r-1)x + 2 = (x + 2)(x^2 + (2r - 1)x + 1) = 0$

For the remaining quadratic equation $x^2 + (2r - 1)x + 1 = 0$ to have real roots, its discriminant must be non-negative: $D = (2r - 1)^2 - 4(1)(1) \ge 0$ $(2r - 1)^2 \ge 4 \implies |2r - 1| \ge 2$ This gives two cases:

1. $2r - 1 \ge 2 \implies 2r \ge 3 \implies r \ge 1.5$
2. $2r - 1 \le -2 \implies 2r \le -1 \implies r \le -0.5$

We need the minimum possible non-negative integer value of $r$. For non-negative integers ($r \ge 0$), $r \ge 1.5$ gives the smallest integer $r = 2$.