\( x^3 + (2r+1)x^2 + (4r − 1)x + 2 = 0 \)
Given that \( -2 \) is a root, the cubic equation factors to:
\((x+2)(x^2 + (2r - 1)x + 1) = 0\)
For the remaining two roots to be real, the quadratic \( x^2 + (2r - 1)x + 1 = 0 \) must possess real roots.
This necessitates a non-negative discriminant:
\( (2r - 1)^2 - 4 \geq 0 \)
\( (2r - 1)^2 \geq 4 \)
Taking the square root of both sides yields:
\( |2r - 1| \geq 2 \)
This implies two conditions: \( 2r - 1 \geq 2 \), leading to \( r \geq \frac{3}{2} \), or \( 2r - 1 \leq -2 \), leading to \( r \leq -\frac{1}{2} \).
Consequently, the smallest possible non-negative integral value for \( r \) is 2.