If one of the solutions to the cubic equation \( x^{3} + 4x^{2} + bx + c = 0 \) is \( -1 \), find values for b and c given that the polynomial has a double root.

To determine the values of \( b \) and \( c \) in the cubic equation \( x^{3} + 4x^{2} + bx + c = 0 \) with a known double root and one root being \( -1 \), let’s denote the double root by \( r \). Therefore, the roots of the cubic can be \( -1, r, r \). Using Vieta's formulas, the sum of the roots gives us: \[ -1 + r + r = -4 \] which simplifies to: \[ 2r - 1 = -4 \implies 2r = -3 \implies r = -\frac{3}{2}. \] Now we have the roots \( -1, -\frac{3}{2}, -\frac{3}{2} \). Now we can express the cubic equation using its roots: \[ (x + 1)\left(x + \frac{3}{2}\right)\left(x + \frac{3}{2}\right) = 0. \] This can be expanded as follows: \[ (x + 1)(x + \frac{3}{2})^2 = (x + 1)\left(x^2 + 3x + \frac{9}{4}\right). \] Next, we multiply it out: \[ = x^3 + 3x^2 + \frac{9}{4}x + x^2 + 3x + \frac{9}{4} = x^3 + 4x^2 + \left(3 + \frac{9}{4}\right)x + \frac{9}{4}. \] Now we combine terms: \[ = x^3 + 4x^2 + \frac{21}{4}x + \frac{9}{4}. \] By comparison with the original polynomial \( x^{3} + 4x^{2} + bx + c = 0 \), we identify: \[ b = \frac{21}{4}, \quad c = \frac{9}{4}. \] Thus, the values of \( b \) and \( c \) are: \[ b = \frac{21}{4}, \quad c = \frac{9}{4}. \]