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In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:

$$ \frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2} $$

Multiplying through to clear the fractions I obtained:

$$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)$$

I found $A=\frac{1}{3}$ by letting $x=1$.

Now in the book they let me know that $B=\frac{2}{3}$, $C=-\frac{1}{3}$, $D=-1$ and $E=0$. But I would really like to figure out how I can find the values for $B, C, D, E$.

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After clearing the denominators we have $$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)(*)$$ as you obtained.
Indeed, $A=\frac{1}{3}$ if we let $x=1$.
Differentiate $(*)$ to get that $$4x^3-3x^2+4x-1=\frac{4}{3}x(x^2+2)+B(4x^3-3x^2+4x-2)+C(3x^2-2x+2)+D(2x-1)+E(**)$$ Differentiate the above relation again to get that $$12x^2-6x+4=\frac{4}{3}(3x^2+2)+B(12x^2-6x+4)+C(6x-2)+2D(***)$$ Differentiate again to get that $$24x-6=8x+B(24x-6)+6C(****)$$ Now let $x=\frac{1}{4}$. It follows that $-2=6C$, so $C=-\frac{1}{3}$.
For $x=0$ in $(****)$ we have $-6=-6B-2$, so $B=\frac{2}{3}$.
For $x=0$ in $(***)$ we obtain that $4=6+2D$, so $D=-1$.
Finally, for $x=0$ in $(**)$ we have $-1=1+E$, so $E=0$.

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    $\begingroup$ Thank you, that is a good strategy. Actually in the book they just introduced this way to solve for the constants two pages back, but seems I already forgot about it. Cheers! $\endgroup$ – Max Jun 27 '19 at 23:35
  • $\begingroup$ @Max We can avoid derivatives and do it purely mentally if we use Heaviside cover-up - see my answer and its link. $\endgroup$ – Bill Dubuque Jun 28 '19 at 16:20
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Setting $x=\sqrt 2i$ (so $x^2=-2$) yields $$-4D-E+\sqrt2(-2D+E)i=2+\sqrt 2 i,\quad\text{whence }\begin{cases}\;2D+E=-2,\\-D+E=1, \end{cases}\iff D=-1,\;E=0.$$ Next, setting $x=0$, you get $$2=4A-2C-E, \quad\text{ so }\; C=-\tfrac13.$$ Last, multiply both sides of the decomposition by $x$ and let $x\to+\infty$, obtaining: $$1=A+B,\quad\text{ so }\;B=\tfrac23.$$

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  • $\begingroup$ @Readers: alternatively, instead of evaluating at $\,x = \sqrt{-2}\,$ we can work $\bmod x^2+2,\,$ see my answer and the linked higher-degree Heaviside cover-up method. $\endgroup$ – Bill Dubuque Jun 28 '19 at 16:18
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Since\begin{multline}A(x^2+2)^2+(Bx+C)(x-1)(x^2+2)+(Dx+E)(x-1)=\\=Ax^4+Bx^4-Bx^3+Cx^3+4Ax^2+2Bx^2-Cx^2+Dx^2+\\-2Bx+2Cx-Dx+Ex+4A-2C-E,\end{multline}solve the system$$\left\{\begin{array}{l}A+B=1\\-B+C=-1\\4A+2B-C+D=2\\-2B+2C-D+E=-1\\4A-2C-E=2.\end{array}\right.$$

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$\overbrace{ x^4\!-\!x^3\!+\!2x^2\!-\!x\!+\!2}^{\large \bbox[5px,border:1px solid #0a0]{\!\!(\color{#c00}{x^2})^2-x(\color{#c00}{x^2})+2\color{#c00}{x^2}\!}\color{#0a0}{-x+2}\ \ }\!\!\! = a(x^2\!+\!2)^2\! + (bx\!+\!c)(x^2\!+\!2)(x\!-\!1) + (dx\!+\!e)(x\!-\!1)\,$ via clear denoms

$x=1 \,\Rightarrow\, 3=9a\,\Rightarrow\,\bbox[5px,border:1px solid #c00]{a=1/3}\ \ $ Compare lead coef's $\,\Rightarrow\, 1 = a\!+\!b=1/3+b\iff \bbox[5px,border:1px solid #c00]{b= 2/3}$

$\!\!\!\!\left.\begin{align}\bmod\ x^2\!+2\ \\ {\rm so}\,\ \color{#c00}{x^2\equiv -2}\ &\end{align}\!\!\right\}\! $ $\!\begin{align} \bbox[5px,border:1px solid #0a0]{4\!+\!2x\!-\!4\!}\!\color{#0a0}{-\!x\!+\!2}&\equiv (d\color{#c00}x+e)(\color{#c00}x-1)\\ \iff\ x\!+\!2 &\equiv (e\!-\!d)x\!-\!e\!\color{#c00}{-\!\!2}d\end{align} $ $\!\!\!\iff\!\!\!\!\! \begin{align} e-\,d &=1\\ -e\!-\!2d& =2\end{align}$ $\!\!\!\iff\!\!\!\!\!\begin{align}-3d&=3\\ 3e&=0\end{align}$ $\!\!\iff\!\!\bbox[5px,border:1px solid #c00]{\!\!\!\begin{align}&d=-1\\ &e\ =\ 0\end{align}\!\!}$

$x=0 \,\Rightarrow\, 2 = 4a\!-\!2c\!-\!e = 4/3-2c\iff 2c=-2/3\iff \bbox[5px,border:1px solid #c00]{c=-1/3}$

Remark $ $ The modular calculation is the higher degree Heaviside cover-up method described here.

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Put $x=1$,

$3=9A$ $\implies$ $A=\dfrac13$.

\begin{align*} x^4-x^3+2x^2-x+2 &= \frac13(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)\\ \frac23x^4-x^3+\frac23x^2-x+\frac23&= (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)\\ \frac23x^3-\frac13x^2+\frac13x-\frac23&= (Bx+C)(x^2+2) + Dx+E\\ \end{align*}

By division, $\dfrac23x^3-\dfrac13x^2+\dfrac13x-\dfrac23=(x^2+2)\left(\dfrac23 x-\dfrac13\right)-x$.

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Hint: Just equate the coefficients of the two polynomials (by FTA they have to be equal), and solve the resulting system.

Thus, $\begin {cases} A+B=1\\-B+C=-1\\4A+2B-C+D=2\\-2B+2C-D+E=-1\\4A-2C-E=2\end{cases}$.

To solve, you could row-reduce the following augmented matrix: $\left(\begin {array}{rrrrr|r}1&1&0&0&0&1\\0&-1&1&0&0&-1\\4&2&-1&1&0&2\\0&-2&2&-1&1&-1\\4&0&-2&0&-1&2\end{array}\right) $

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