Partial Fractions

**jgv115** · December 13th 2010, 03:43 PM

I know how to do them when there is only linear factors in the denominator but when it has a repeated linear factor and a single linear factor then I get confused.

In my book it has this example (I will provide a picture since it's too long to type)

I can understand where it splits it up into A,B and C but when I get confused is when they do

$x=1$ and let $x=-1$

Are they just pulling random numbers from no where or something? How does that part work?

**pickslides** · December 13th 2010, 03:48 PM

Originally Posted by jgv115

$x=1$ and let $x=-1$

Are they just pulling random numbers from no where or something? How does that part work?

Any numbers will work to solve a system for A,B,C but notice x=1,-1 will cancel terms and make life easier.

**snowtea** · December 13th 2010, 03:50 PM

Plugging in numbers is a standard way of solving for unknown constants.

For example, if two functions f and g are equal, then f(x) = g(x) for all x.

Now, in this example, consider f(x) = 2x + 10
and g(x) = A(x - 1)^2 + B(x + 1)(x - 1) + C(x + 1)

We know f(x) = g(x).

This means that it has to be the case that f(1) = g(1) and f(-1) = g(-1), which is what the problem does when it sets x = 1 and x = -1.

The truth is, the book could've chosen any values x = 10, x = 20, x = 0.5, x = pi, ... and solved, and the solution for A, B, C will be the same.
But why did they choose x = 1 and x = -1?
Look at the expression for g(x) = A(x - 1)^2 + B(x + 1)(x - 1) + C(x + 1), if you could choose any value of x to solve for A, B, and C, what would you pick?
x = 1 gets rid of A and B completely leaving only C
x = -1 gets rid of B and C completely leaving only A
These are 2 values that are easy to solve with, so the book chooses them.

**Prove It** · December 13th 2010, 03:50 PM

What they are doing is choosing values of $\displaystyle x$ that will make two of the terms $\displaystyle 0$ , thus eliminating two terms and making it possible to solve for the third. Since this is an equation, you can substitute whatever values of $\displaystyle x$ you like as long as you substitute for every $\displaystyle x$ on both sides.

Since you have $\displaystyle A(x - 1)^2$ , to make this $\displaystyle 0$ you need to choose $\displaystyle x = 1$ , and since you have $\displaystyle C(x + 1)$ , to make this $\displaystyle 0$ you need to choose $\displaystyle x = -1$ . Also since you have $\displaystyle B(x - 1)(x + 1)$ , choosing either $\displaystyle x = -1$ or $\displaystyle x = 1$ makes it $\displaystyle 0$ .

**jgv115** · December 13th 2010, 03:51 PM

oh!!!!!!!!!!!! alright

ok for this question:

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

How would I set it up?

Would it be

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

**snowtea** · December 13th 2010, 03:56 PM

If $x = 3$

$2(3) + 10 = A(3 - 1)^2 + B(3 + 1)(3 - 1) + C(3 + 1)$

The A and B disappear only for $x = 1$

The book jumps a bit in this step.

**snowtea** · December 13th 2010, 04:03 PM

Yes, you setup the problem as:

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

**jgv115** · December 13th 2010, 04:07 PM

NICE!! I GOT IT

If I take the common denominator I get

$2x+3 = A(x-3)+B$

Then let x =3 so B=9

Then I just let x = 1 and sub B in and I got the right answer!

Yay thanks guys

I'm so happy

**dd86** · December 13th 2010, 04:11 PM

When both your LHS and RHS of the equations still have denominators, you can't choose a value that will make any of your denominators zero i.e x=3 when all your denominators are x-3. This would then lead to an undefined value.

If you were in a rush and careless, you might end up leaving those terms out and this could lead to serious miscalculations of your end-result.

**snowtea** · December 13th 2010, 04:23 PM

Yes, what dd86 says is true. The reason you can multiply and then use values for x that would've made denominators zero is quite subtle.

Instead of getting into the details, let me just say:
It is always a good idea to check your answers by plugging them back into your original problem. Even if you are unsure about the correctness of a method, you can have confidence in the answer by checking it with the original problem.

**pickslides** · December 13th 2010, 06:36 PM

Originally Posted by jgv115

Would it be

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

Yep..

**jgv115** · December 13th 2010, 06:44 PM

mm.. anyone care to explain why you have to write it like that? Or is it just a rule?

**pickslides** · December 13th 2010, 06:49 PM

It is the rule for repeated linear factors.

If you choose otherwise the equation won't balance.

**dd86** · December 13th 2010, 06:52 PM

Not sure whether it's a rule. It's more like a process to me. A process that saves time and gives you and idea how to solve it.

But the numerators do differ depending on what your denominator is. So you have to be careful with that.

Why don't you try reading up on the cover-up rule for partial fractions? It might help you save some time also. However, I'm not sure whether the examiner in your school would accept it if you used it in an exam...

**Prove It** · December 13th 2010, 06:53 PM

Originally Posted by jgv115

mm.. anyone care to explain why you have to write it like that? Or is it just a rule?

You might expect to write it as $\displaystyle \frac{A}{x - 3} + \frac{B}{x - 3}$ . But notice there is a common denominator, so this simplifies to $\displaystyle \frac{A+B}{x-3}$ , which does not have the required denominator $\displaystyle (x-3)^2$ .

In order to get the required denominator, you need to write $\displaystyle \frac{A}{x - 3} + \frac{B}{(x - 3)^2}$ or $\displaystyle \frac{Ax+B}{(x-3)^2}$ . The easiest method is to have the numerator be a polynomial of one less degree than the denominator.

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