# Calculating Partial Fractions find A, B and C

1. Jun 7, 2014

### lubo

1. The problem statement, all variables and given/known data

For the equation shown below:

x2+2x+3 / (x2+9)(X-3) = Ax+B/(x2+9) + C/(x-3)

Find A, B and C

2. Relevant equations

3. The attempt at a solution

C = 1
B = 2
A = ?

Find C which = 1 by putting x=3 and working out x2+2x+3/(x2+9),

then multiply out equation to find A and B

x2+2x+3 = C(X2+9) + (Ax+B)/(X-3)

Combine terms to give:

x2+9+Ax2-3Ax+Bx-3B

= x2(1+A)+ x(B-3A) + 9 -3B

if coefficients = 0 then x2 = 0 then A = -1
9-3B = 3 therefore B = 2

B-3A = 0 therefore A = 2/3 but I think this answer is wrong? I now have 2 answers for A?

2. Jun 7, 2014

### micromass

Ok, so what do you mean exactly? The way you wrote it is

$$x^2 + 2x + \frac{3}{x^2 + 9}(x-3) = Ax + \frac{B}{x^2 +9} + \frac{C}{x-3}$$

Is this what you meant?

3. Jun 7, 2014

### lubo

I have changed it to reflect what you mean, thanks.

4. Jun 7, 2014

### HallsofIvy

I presume you mean
$$\frac{Ax+ B}{x^2+ 9}+ \frac{C}{x- 3}$$

I take it you meant that you first multiplied both sides by x- 3 so you had
$$\frac{x^2+ 2x+ 3}{x^2+ 9}= \frac{Ax+ B}{x^2+ 9}(x- 3)+ C$$
Taking x= 3, this is $$\frac{9+ 6+ 3}{9+ 9}= 1= C$$.

On the right side you mean? And (Ax+ B)(x- 3), not (Ax+ B)/(x- 3).

The way you have it set up the coefficients are not equal to 0- they are equal to 1, 2, and 3, the coefficients on the left side of the equation. A+ 1= 1 gives A= 0. 9- 3B= 3 gives B= 2.
With A= 0, B= 2, B- 3A= 2 automatically.

5. Jun 7, 2014

### micromass

Thank you!

I agree until here, but

The coefficients shouldn't be equal to $0$. What you have is

$$x^2 + 2x + 3 = (1+A)x^2 + (B-3A)x + (9 - 3B)$$

6. Jun 7, 2014

### Ray Vickson

Do not make any assumptions about A,B and C: just write your function f(x) as
$$f(x) =\frac{x^2+2x+3}{(x^2+9)(x-3)} = \frac{Ax+B}{x^2+9}+\frac{C}{x-3} = \frac{(A+C)x^2 +(B-3A)x + (9C-3B)}{(x^2+9)(x-3)}$$
and so get three equations for the three parameters A,B,C. Solve them and see what you get.

7. Jun 7, 2014

### lubo

micromass

HallsofIvy

How did you find the Coefficients as I thought you could automatically say that they are equal to Zero?

8. Jun 7, 2014

### micromass

You have found the correct equation

$$x^2+2x+3 = C(x^2+9) + (Ax+B)(x-3)$$

which must be satisfied for all $x$. In particular, it must be satisfied for $x=3$. Using that, we get $C=1$. Thus we have that the following equation

$$x^2+2x+3 = x^2+9 + (Ax+B)(x-3)$$

must be satisfied for all $x$. So what you did is work out the right-hand side, to get

$$x^2 + 2x + 3 = x^2 + 9 + Ax^2 - 3Ax + Bx - 3B$$

which is

$$x^2 + 2x + 3 = (1 + A)x^2 + (B - 3A)x + (9-3B)$$

which again must be satisfied for all $x$. Thus we get the equations

$$1 = 1+A,~2 = B-3A,~3=9-3B$$

9. Jun 7, 2014

### lubo

Why should 1 = 1+A

Why 2 = B-3A

Why 3=9-3B

I found 9-3B = 3 because I put x = 0 in (x^2 + 2x + 3)

This is what I am failing to understand i.e. why choose these and not just Zero?

Thanks again as I realise this must be obvious to you.

10. Jun 7, 2014

### micromass

OK, that is a good question. So assume that there exist $A$ and $B$ such that

$$x^2+2x+3=(1+A)x^2+(B−3A)x+(9−3B)~~~~(1)$$

are true all $x$. In particular, it must be satisfied for $x=0$. This gets us the equation

$$3 = 9-3B~~~~(2)$$

But it must also be true for $x=1$, which gets us

$$1^2 + 2\cdot 1 + 3 = (1+A) + (B-3A) + (9 - 3B)~~~~(3)$$

And it must also be true for $x=-1$, which gets us

$$(-1)^2 + 2\cdot (-1) + 3 = (1+A) -(B-3A) +(9 - 3B)~~~~(4)$$

Subtracting $(2)$ from $(3)$ and $(4)$ leaves us with the two equations

$$1 + 2 = (1+A) + (B-3A)~~~~(3^\prime)$$

and

$$1 - 2 = (1+A) - (B-3A)~~~~(4^\prime)$$

Adding up these two gets us

$$2 = 2(1+A)$$

or just $(1+A) = 1$ which I call $(5)$. Subtracting $(5)$ from $(3^\prime)$ then gets us

$$2 = B-3A$$

So we finally see that we have the equations

$$1+A = 1,~2= B-3A,~3=9-3B$$

Of course, I didn't do all that just to find these equations. What I said was simply that since

$$x^2+2x+3=(1+A)x^2+(B−3A)x+(9−3B)$$

must be true for all $x$, then all the coefficients must equal. So the coefficients of $x^2$ must equal, which gets us $1+A=1$. The coefficients of $x$ must equal, which gets us $2 = B-3A$, and the constant terms must equal, so $3 = 9-3B$.
But then you can ask why must the coefficients equal? Well, my above (long) arguments proves it. In general, whenever you are in the situation that

$$Ax^2 + Bx + C = \alpha x^2 + \beta x + \gamma$$

you can apply my argument above and get that $A=\alpha$, $B=\beta$ and $C=\gamma$. So all coefficients equal.

11. Jun 7, 2014

### LCKurtz

Or, when you have $f(x)\equiv g(x)$ you can use that $f(0) = g(0), ~f'(0) = g'(0),~ f''(0) = g''(0)$ to get the constants equal in this example.

12. Jun 7, 2014

### micromass

Yes, but this is in the precalculus forum, so I didn't want to use the easier way of derivatives.

13. Jun 7, 2014

### LCKurtz

I assumed that. But since they post all kinds of calculus and beyond stuff in that forum anyway, and students don't usually encounter partial fractions for anything before calculus, I'm guessing the OP is in a calculus course and might appreciate the derivative method.

14. Jun 7, 2014

### SammyS

Staff Emeritus
Equating coefficients (of any particular power of x) is a method that's often used per-calculus.

Two polynomials are equivalent only if their coefficients "match up" .