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:

Question

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:

Accepted Answer

Given that $$ -2 $$ is a root of the cubic equation:  
$$ x^3 + (2r + 1)x^2 + (4r - 1)x + 2 = 0 $$,  
the equation can be factored as:  
$$ (x + 2)(x^2 + (2r - 1)x + 1) = 0 $$.

For the other two roots to be real, the discriminant of the quadratic factor  
$$ x^2 + (2r - 1)x + 1 $$  
must be non-negative. The discriminant is given by:  
$$ \Delta = b^2 - 4ac $$, where $$ a = 1, b = (2r - 1), c = 1 $$.  
Thus:  
$$ \Delta = (2r - 1)^2 - 4(1)(1) $$.  
Simplify:  
$$ \Delta = (2r - 1)^2 - 4 \geq 0 $$.

Expanding and simplifying:  
$$ (2r - 1)^2 \geq 4 $$.  
$$ |2r - 1| \geq 2 $$.

Solving for $$ r $$:  
$$ 2r - 1 \geq 2 \quad 	ext{or} \quad -(2r - 1) \geq 2 $$.  
From the first inequality:  
$$ r \geq \frac{3}{2} $$.  
From the second inequality:  
$$ r \leq -\frac{1}{2} $$.

Since we are looking for the minimum non-negative integer value of $$ r $$, the smallest possible value is **$$ r = 2 $$**.