Question 2.4.5: Find constants A, B, C, D, E, and F such that 3x^5 - 2x^4 - ...

We first multiply both sides of the equation by (x^2 + 1)^2 (x^2 + x + 1) to get

3x^5 – 2x^4 – x + 1 = (Ax + B)(x^2 + 1)(x^2 + x + 1) + (Cx + D)(x^2 + x + 1)  \\ + (Ex + F)(x^2 + 1)^2

Expanding the right and collecting coefficients of like powers of x gives

3x^5 – 2x^4 – x + 1 = (A + E)x^5 + (A + B + F)x^4 + (2A + B + C + 2E)x^3 \\

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + (A + 2B + C + D + 2F)x^2 + (A + B + C + D + E)x + B + D + F

Comparing coefficients of like powers of x gives the system of linear equations

A + E = 3 \\

A + B + F = -2 \\

2A + B + C +2E = 0 \\

A + 2B + C + D + 2F = 0 \\

  A + B + C + D + E = -1 \\

B + D + F = 1

Row reducing the corresponding augmented matrix gives

\left [ \begin{matrix} 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 0 & 2 & 0 \\ 1 & 2 & 1 & 1 & 0 & 2 \\ 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \end{matrix} \left | \begin{matrix} 3 \\ -2 \\ 0 \\ 0 \\ -1 \\ 1 \end{matrix} \right.\right] \thicksim \left [ \begin{matrix} 1 & 0 & 0 & 0` & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} \left | \begin{matrix} -1 \\ -7 \\ 1 \\ 2 \\ 4 \\ 6 \end{matrix} \right.\right ]

Hence,

\frac{3x^5 – 2x^4 – x + 1}{(x^2 + 1)^2 (x^2 + x + 1)} = \frac{-x – 7}{x^2 + 1} + \frac{x + 2}{(x^2 + 1)^2} + \frac{4x + 6}{x^2 + x + 1}