# Finding constants in partial fraction

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 $$B, C, D, E$$.

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 $$E=0$$.

• 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! – Max Jun 27 '19 at 23:35
• @Max We can avoid derivatives and do it purely mentally if we use Heaviside cover-up - see my answer and its link. – Bill Dubuque Jun 28 '19 at 16:20

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: $$1=A+B,\quad\text{ so }\;B=\tfrac23.$$

• @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. – Bill Dubuque Jun 28 '19 at 16:18

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$$\left\{\begin{array}{l}A+B=1\\-B+C=-1\\4A+2B-C+D=2\\-2B+2C-D+E=-1\\4A-2C-E=2.\end{array}\right.$$

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

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

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: $$\left(\begin {array}{rrrrr|r}1&1&0&0&0&1\\0&-1&1&0&0&-1\\4&2&-1&1&0&2\\0&-2&2&-1&1&-1\\4&0&-2&0&-1&2\end{array}\right)$$