Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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;

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

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.

share|improve this question
    
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

3 Answers 3

up vote 5 down vote accepted

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 numerators:

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

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

share|improve this answer
    
I was not here for hours. In fact, my int line is expired today. Glad to see you on here. :+) –  Babak S. Jul 30 '13 at 0:23
    
Hello, @Babak! I have not done too well today, at math.se. Good to see you: that makes my day complete! –  amWhy Jul 30 '13 at 0:27
    
I have to be off here and on for next hours till my line gets charged again. This is a temporary access. –  Babak S. Jul 30 '13 at 0:31
    
@Amzoti Thanks. Yes, it was a lot of typing, but not comparable, in typing, to some of your thorough answers: you've got some gems! –  amWhy Jul 30 '13 at 1:12
    
Thank you so much @amWhy! This really helped. –  Sugi Jul 30 '13 at 4:17

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.

share|improve this answer

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 $$A=-\frac{1}{2}\quad,\quad B=\frac{1}{2}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.