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

We need to find the value of $r$ when one root is $-2$ and the other two roots are real.

Let us work through this using Vieta's formulas - these connect the coefficients of a polynomial to sums and products of its roots.

---

For the cubic $x^3 + (2r + 1)x^2 + (4r - 1)x + 2 = 0$, if the three roots are $-2$, $\alpha$, and $\beta$:

Product of all roots: $(-2) \cdot \alpha \cdot \beta = -\frac{2}{1} = -2$

This gives us: $\alpha \cdot \beta = 1$

Sum of all roots: $(-2) + \alpha + \beta = -\frac{(2r + 1)}{1} = -(2r + 1)$

This gives us: $\alpha + \beta = -2r - 1 + 2 = -2r + 1$

---

Since $\alpha$ and $\beta$ are real numbers with $\alpha \cdot \beta = 1$, we need to find when this is possible.

Key insight: When two real numbers have a positive product, they're either both positive or both negative.

If both are positive: Using AM-GM inequality, $\alpha + \beta \geq 2\sqrt{\alpha\beta} = 2\sqrt{1} = 2$

If both are negative: Then $\alpha + \beta \leq -2\sqrt{|\alpha||\beta|} = -2\sqrt{1} = -2$

Therefore: $\alpha + \beta \geq 2$ or $\alpha + \beta \leq -2$

---

Substituting $\alpha + \beta = -2r + 1$:

Case 1: $-2r + 1 \geq 2$

$-2r \geq 1$

$r \leq -\frac{1}{2}$

Case 2: $-2r + 1 \leq -2$

$-2r \leq -3$

$r \geq \frac{3}{2}$

So: $r \leq -\frac{1}{2}$ or $r \geq \frac{3}{2}$

---

Since we need non-negative integer values, only $r \geq \frac{3}{2}$ is relevant.

The minimum non-negative integer satisfying $r \geq \frac{3}{2}$ is $r = 2$.

---

Let us check $r = 2$:

$\alpha + \beta = -2(2) + 1 = -3$

$\alpha \beta = 1$

The quadratic with roots $\alpha, \beta$ is: $t^2 - (-3)t + 1 = t^2 + 3t + 1 = 0$

Discriminant $= (-3)^2 - 4(1)(1) = 9 - 4 = 5 > 0$

Since the discriminant is positive, both roots are real.

Therefore, the minimum possible non-negative integer value of $r$ is $\boxed{2}$.