Partial fractions- Equating coefficients

This is a partial fraction question. Kindly could you assist me with finding the values for A, B and C as I'm not very clear on how to go about it. The question is as follows;

$$\frac{3-x}{(x^2+3)(x+3)}$$

And we want A, B, C so that $$\frac{3-x}{(x^2+3)(x+3)} = \frac{Ax+B}{x^2 + 3} + \frac C{x+3}$$

I know that $A+B= 3$, $3B+C= -1$ and $3A+3C= 0$ The problem I'm facing here is with equating coefficients. I also know that after equating coefficients, the final values you get for A,B and C are as follows;

\begin{align} A&= -1/2\\ B&= 1/2\\ C&= 1/2 \end{align}

But how do you solve with equating coefficients? Can you provide me step-by-step working for this please. I'm not facing any difficulty with partial fractions but just equating coefficients part in this case. Please help.

Many thanks.

• First, your expression isn't clear. Use more parentheses, or much better: write mathematics with LaTeX . Second, if you already know that $\,A+B=3\;$, then it can't be $\,A=-\frac12=-B\;$ ... – DonAntonio Jul 29 '13 at 11:57

It looks like you already have equated coefficients:

$$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac C{x+3}$$

$$\iff \frac{\color{blue}{(Ax+B)(x+3)+C(x^2+3)}}{(x^2+3)(x+3)}= \frac{\color{blue}{3-x}}{(x^2+3)(x+3)}$$ Equating the numerators:

$$(Ax+B)(x+3)+C(x^2+3) = 0\cdot x^2 -x + 3\tag{1}$$ Expanding the LHS of equation $$(1)$$, gathering like terms:

$$Ax^2 + (B + 3A)x + 3B + Cx^2 + 3C = 0x^2 - x + 3$$ $$\iff \color{blue}{\bf (A + C)}x^2 +\color{red}{\bf (3A + B)}x + \color{green}{\bf (3B + 3C)} = \color{blue}{\bf 0}x^2 + \color{red}{\bf (-1)}x + \color{green}{\bf 3}\tag{2}$$

Match up (color coded above) coefficients on LHS with those on RHS of $$(2)$$:

$$\iff \color{blue }{\bf A + C = 0}, \quad \color{red}{\bf 3A + B = -1}, \quad \color{green}{\bf 3(B + C) = 3 \iff B+C = 1}$$

Indeed, this gives us a system of $$\bf 3$$ equations in $$\bf 3$$ unknowns, from which we can solve for the "unknowns" $$A, B, \;\text{and}\; C$$.

\begin{align} A + C & = 0 \tag{i}\\ 3A + B & = -1 \tag{ii}\\ B+ C & = 1\tag{iii}\end{align}

Subtract $$(iii)$$ from $$(i)$$: $$A - B = -1\tag{iv}$$

Adding $$(iv)$$ to $$(ii)$$ gives us $$4A = -2 \iff A = -\dfrac 12\tag{A}$$ From $$(i)$$: $$A = -\dfrac 12 \implies C = \dfrac 12\tag{C}$$ From $$(iii)$$: $$C = \dfrac 12 \implies B = \dfrac 12\tag{B}$$

Therefore, we have the following function, replacing coefficients A, B, C with their found values:

$$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac C{x+3}$$ $$= \frac{-\frac12\cdot x+\frac 12}{x^2+3}+\frac {\frac 12}{x+3}=\dfrac 12\left(\frac{1-x}{x^2 + 3}\right) + \dfrac 12\left(\frac 1{x+3}\right)$$

• Thank you so much @amWhy! This really helped. – Sugi Jul 30 '13 at 4:17
• You're welcome! Thank you ;-) – Namaste Jul 31 '13 at 10:11

Here I give an alternative method:

Let denote the partial fraction by

$$F(x)=\frac{3-x}{(x^2+3)(x+3)}$$ I guess from your question that you want a decompostion of $F$ over $\mathbb R$ so we have:

$$F(x)=\frac{C}{x+3}+\frac{Ax+B}{x^2+3}$$

We find $C$ by: $$C=(x+3)F(x)\big|_{x=-3}=\frac{1}{2}$$ and for $A$ and $B$ we have $$F(x)(x^2+3)\big|_{x=i\sqrt{3}}=\frac{3-i\sqrt{3}}{i\sqrt{3}+3}=\frac{1}{12}(6-i6\sqrt{3})=Ai\sqrt{3}+B$$ so we find $$A=-\frac{1}{2}\quad,\quad B=\frac{1}{2}$$

Step 1: Write $$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac{C}{x+3}$$ Step 2: Expand the RHS to get $$\frac{Ax+B}{x^2+3}+\frac{C}{x+3}=\frac{(Ax+B)(x+3)+C(x^2+3)}{(x^2+3)(x+3)}=\frac{(A+C)x^2+(3A+B)x+3(B+C)}{(x^2+3)(x+3)}$$ Step 3: Compare the coefficients. $$x^2:A+C=0,\quad x:(3A+B)=-1,\quad\text{constant}: 3(B+C)=3$$ Step 4: Solve simultaneous equations.