Solving Partial Fractions

[david7]

Hello

Can someone help me please I am really struggling with partial fractions I have 2 questions to complete and would appreciate some help, my questions are:
x2-3x+6/x(x-2)(x-1) and x2+9x+8/x2+x-6 I know there are some clever people on here if anyone can help with the steps aswell please.


Thanks in advance David

[HowardKennitby]

this might help

[Buenchico]

Start by writing the expression in the form you require (introducing unknowns A, B & C).

Then multiply the numerator and denominator of each fraction accordingly, in order to create a common denominator.

Then group like terms (from the numerators) together. i.e. collect up x-squared, x and numeric terms.

Next, compare the coefficients of each term, to create three simultaneous linear equations.

Solve those equations.

Job done (but you should 'multiply out' your answer, in order to check it).

Here's the first one done for you:
http://i40.tinypic.com/2utlv10.jpg

Chris

[david7]

Thank you very much I appreciate your help ;-)

[factor-fiction]

To start you off on the second one:

The denominator x² + x – 6 factorises into (x+3)(x-2)

You may also want to factorise the numerator x² +9x +8 into (x+8)(x+1)

Then proceed using the above method explained by Chris of comparing coefficients

[david7]

Hello Factor30 & Buenchico

Could you solve the second question for me please sorry for being cheecky I really struggle with partial fractions and am trying hard to understand them.

Kind regards

David ;-)

[HowardKennitby]

Has anyone ever found a situation where things like this are useful?

[bibblebub]

yes i found this sort of thing useful when i worked at the National Physical Laboratory for a short while and had to calculate the intensity contours for an image generated by an x-ray telescope (required manipulating Jacobians if you're interested)

[HowardKennitby]

That's something to do with matrices isn't it?

[bibblebub]

certainly is

[HowardKennitby]

believe it or not I actually studied that sort of thing. Long time ago now and all memory of it has faded.

[bibblebub]

same here, the last time i did any real maths was 30 years ago but can still dredge up some of the basics if required

[TheNovice]

just watched that video....how the hell did I get a GCE O'level in additional maths. Must have understood it then, damned if I do now!!!!!!

[factor-fiction]

David- is the second question typed correctly?
Is it :
x² +9x +8
_________
x² + x – 6

You may also want to factorise the numerator x² +9x +8 into (x+8)(x+1)

[david7]

Hello

Yes the question is typed correctly how do you get (x+1)

[factor-fiction]

Sorry- you don't need to bother factorising the numerator.

I will start you off with what i think is the right method.

First, factorise the denominator into (x+3)(x-2).

x² +9x +8
_________
(x+3)(x-2)

Set this as Ξ to A/(x+3) + B/(x-2) + C

Proceeding in the way Chris did, by putting every term over (x+3)(x-2):

x² +9x +8 Ξ A(x-2) +B(x+3) + C(x+3)(x-2)

Expanding:
x² +9x +8 Ξ Ax-2A +Bx +3B +Cx² + Cx - 6C

Comparing coefficients:
(of x²).............C=1
(of x)..... ... A+B+C=9, so A+B=8
(of constants)...8=-2A +3B-6C = -2A+3B-6, so 14= 3B-2A

Solving these simultaneous equations gives, I think, A=2, B=6 and C=1

Put these values now into
A/(x+3) + B/(x-2) + C

It's a bit late at night for this but I hope you can follow the method and check my working. Also, I suggest you try substituting in some values of x and make sure it works

[factor-fiction]

Yes, I tried x=1 into the original expression and got -4.5

Then I substituted into my
2/(x+3) + 6/(x-2) + 1
and also got -4.5

[vascop]

One of the most useful applications of partial fractions is in integration.
The integral of A/(Bx+C) with respect to x is (A/B)log(Bx+C) and so the integration of something like x2-3x+6/x(x-2)(x-1) can be reduced to a sum of integrals like A/(Bx+C) which are easy to do.

[factor-fiction]

Did this help, david7?

[vascop]

@factor
I like your solution. Another way which saves a bit of work is to start from your equation:
x² +9x +8 Ξ A(x-2) +B(x+3) + C(x+3)(x-2)
and instead of multiplying this out, notice that if x=2 or 3 that the second and third terms are zero. So put x=2 in the above gives:
4+18+8=5B, so B=6
Putting x=-3 gives:
9-27+8=-5A, so A=2
and then putting x=0 and substituting for A and B gives:
8=-4+18-6C, so C=1