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Enter (A,B,C,D,E) in order below if A, B, C, D, and E are the coefficients of the partial fractions expansion of $$f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = Ax^2 + Bx + C + \frac{D}{x+2} + \frac{E}{x-1}.$$

Mar 9, 2018

#1
+24983
0

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

$$\displaystyle f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = Ax^2 + Bx + C + \frac{D}{x+2} + \frac{E}{x-1}.$$

f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = Ax^2 + Bx + C + \frac{D}{x+2} + \frac{E}{x-1}.

$$\small{ \begin{array}{|rcll|} \hline 3\cdot\dfrac{x^4+x^3+x^2+1}{x^2+x-2} &=& 3\cdot\dfrac{x^4+x^3+x^2+1}{(x+2)(x-1)} \\ \\ \hline \\ 3\cdot\dfrac{x^4+x^3+x^2+1}{(x+2)(x-1)} &=& Ax^2 + Bx + C + \dfrac{D}{x+2} + \dfrac{E}{x-1} \qquad | \qquad \cdot (x+2)(x-1) \\ \hline \end{array} }$$

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

$$\small{ \begin{array}{lrcll} \hline \mathbf{x = 1:} & 3(1 +1+1+1) &=& A\cdot 0 + B\cdot 0 + C\cdot 0 + D\cdot 0 + E\cdot (1+2) \\ & 12 &=& 3E \\ & \mathbf{E} &\mathbf{=}& \mathbf{4} \\ \\ \hline \mathbf{x = -2:} & 3(16-8+4+1) &=& A\cdot 0 + B\cdot 0 + C\cdot 0 + D\cdot (-2-1) + E\cdot 0 \\ & 3\cdot 13 &=& -3D \\ & 13 &=& - D \\ & \mathbf{D} &\mathbf{=}& \mathbf{-13} \\ \\ \hline \mathbf{x = 0:} & 3(0+0+0+1) &=& A\cdot 0 + B\cdot 0 + C\cdot(0+2)(0-1) + D\cdot (0-1) + E\cdot (0+2) \\ & 3 &=& -2C-D+2E \\ & 3 &=& -2C-(-13)+2\cdot 4 \\ & 3 &=& -2C+13+8 \\ & 3 &=& -2C+ 21 \\ & 2C &=& 21 -3 \\ & 2C &=& 18 \\ & \mathbf{C} &\mathbf{=}& \mathbf{9} \\ \\ \hline \mathbf{x = -1:} & 3(1 -1+1+1) &=& A\cdot 1(-1+2)(-1-1) + B\cdot (-1)(-1+2)(-1-1) \\ & &+& C\cdot (-1+2)(-1-1) + D\cdot (-1-1) + E\cdot (-1+2) \\ & 6 &=& -2A+2B-2C-2D+E \\ & 6 &=& -2A+2B-2\cdot 9 +2\cdot 13+ 4 \qquad | \qquad : 2 \\ & 3 &=& -A+B-9 +13+ 2 \\ & 3 &=& -A+B+6 \\ & \mathbf{-A+B} &\mathbf{=}& \mathbf{-3} \qquad (1) \\ \\ \hline \mathbf{x = 2:} & 3(16+8+4+1) &=& A\cdot 4(2+2)(2-1) + B\cdot 2(2+2)(2-1) \\ & &+& C\cdot (2+2)(2-1) + D\cdot (2-1) + E\cdot (2+2) \\ & 87 &=& 16A+8B+4C+D+4E \\ & 87 &=& 16A+8B+4\cdot 9 -13+4\cdot 4 \\ & 87 &=& 16A+8B+36 -13+16 \\ & 87 &=& 16A+8B+39 \\ & 16A+8B&=& 87 - 39 \\ & 16A+8B&=& 48 \qquad | \qquad : 8 \\ & \mathbf{2A+B} &\mathbf{=}& \mathbf{6} \qquad (2) \\ \hline \end{array} }$$

$$\begin{array}{|lrclcrclr|} \hline (2) & 2A+B &=& 6 &\qquad & -A+B &=& -3 & (1) \\ & & & &\qquad & B &=& -3+A \\ & 2A-3+A &=& 6 \\ & 3A &=& 9 \\ & \mathbf{A } &\mathbf{=}&\mathbf{ 3} \\ & & & &\qquad & B &=& -3+3 \\ & & & &\qquad & \mathbf{B} &\mathbf{=}&\mathbf{ 0} \\ \hline \end{array}$$

$$\displaystyle f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = 3x^2 + 0\cdot x + 9 - \frac{13}{x+2} + \frac{4}{x-1}$$

$$\mathbf{(A,B,C,D,E) = (3,0,9,-13,4)}$$

Mar 9, 2018