Enter (A,B,C,D) in order below if A, B, C, and D are the coefficients of the partial fractions expansion of ​

Enter (A,B,C,D) in order below if A, B, C, and D are the coefficients of the partial fractions expansion of

\(\displaystyle12\cdot\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}\)

12\cdot\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}

\(\small{ \begin{array}{|rcll|} \hline 12\cdot\dfrac{x^3+4}{(x^2-1)(x^2+3x+2)} &=& 12\cdot\dfrac{x^3+4}{(x-1)(x+1)(x+2)(x+1)} \\ \\ \hline \\ 12\cdot\frac{x^3+4}{(x-1)(x+1)(x+2)(x+1)} &=& \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2} \qquad | \qquad \cdot (x-1)(x+2)(x+1)^2 \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline 12(x^3+4) &=& A\cdot (x+2)(x+1)^2 + B\cdot (x-1)(x+2)^2 \\ &+& C\cdot (x-1)(x+2)(x+1) + D\cdot (x-1)(x+2) \\ \hline \end{array}\)

\(\small{ \begin{array}{lrcll} \hline \mathbf{x = 1:} & 12(1 + 4) &=& A\cdot (1+2)(1+1)^2 + B\cdot 0 + C\cdot 0 + D\cdot 0 \\ & 60 &=& A\cdot 3 \cdot 4 \\ & 60 &=& 12A \\ & \mathbf{A} &\mathbf{=}& \mathbf{5} \\ \\ \hline \mathbf{x = -1:} & 12(-1 + 4) &=& A\cdot 0 + B\cdot 0 + C\cdot 0 + D\cdot (-1-1)(-1+2) \\ & 36 &=& -2D \\ & \mathbf{D} &\mathbf{=}& \mathbf{-18} \\ \\ \hline \mathbf{x = -2:} & 12(-8 + 4) &=& A\cdot 0 + B\cdot (-2-1)(-2+1)^2 + C\cdot 0 + D\cdot 0 \\ & -48 &=& -3B \\ & 48 &=& 3B \\ & \mathbf{B} &\mathbf{=}& \mathbf{16} \\ \\ \hline \mathbf{x =0:} & 12(0 + 4) &=& A\cdot 2 + B\cdot (0-1)(0+1)^2 + C\cdot (0-1)(0+2)(0+1)\\ & &+& D\cdot (0-1)(0+2) \\ & 48 &=& 2A-B-2C-2D \\ & 48 &=& 2\cdot 5-16-2C-2\cdot(-18) \\ & 48 &=& 10-16-2C+36 \\ & 48 &=& 30-2C \\ & 2C &=& 30-48 \\ & 2C &=& -18 \\ & \mathbf{C} &\mathbf{=}& \mathbf{-9} \\ \hline \end{array} }\)

\(\displaystyle 12\cdot\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{5}{x-1} + \frac{16}{x+2} - \frac{9}{x+1} - \frac{18}{(x+1)^2}\)

\(\mathbf{(A,B,C,D) = (5,16,-9,-18)}\)