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# C4 Partial fractions watch

1. Cannot do part a) example in book is awful and wasnt shown how to do it with 3 factors. Can anyone help?
2. (Original post by AndyOC)
Cannot do part a) example in book is awful and wasnt shown how to do it with 3 factors. Can anyone help?
As far as ive got with it
3. (Original post by AndyOC)
Cannot do part a) example in book is awful and wasnt shown how to do it with 3 factors. Can anyone help?
If any factors aren't repeated then I believe that you just add C/third factor to how you would normally do it.
4. It's the same as you would normally do it, except with a third unknown - typically 'C'.
Also your working is wrong, its just A B C, one for each denominator factor.
From there you could use your own method of getting the simultaneous equations.
5. (Original post by AndyOC)
As far as ive got with it
You don't need the A from the start, but the B, C and D are fine.
6. the A is superfluous
7. (Original post by SeanFM)
You don't need the A from the start, but the B, C and D are fine.
(Original post by the bear)
the A is superfluous
why is it unnecessary?
8. (Original post by AndyOC)
why is it unnecessary?
since the order of the top of your fraction is less than the underneath you do not need any extra bits.
9. (Original post by AndyOC)
why is it unnecessary?
You have three linear factors and the original numerator is of order 3 or less. So you can split it into three separate fractions each with a constant numerator.
10. (Original post by the bear)
since the order of the top of your fraction is less than the underneath you do not need any extra bits.
(Original post by Muttley79)
You have three linear factors and the original numerator is of order 3 or less. So you can split it into three separate fractions each with a constant numerator.
Ok but i still don't know where to go from there
11. (Original post by AndyOC)
Ok but i still don't know where to go from there
OK - get rid of the A - then write the right hand side over the common denominator.

B (x + 2)(x + 1) + C (x + 3)(x + 1) + D(x + 3)(x + 2) will be the numerator.

Can you go from here?
12. (Original post by Muttley79)
OK - get rid of the A - then write the right hand side over the common denominator.

B (x + 2)(x + 1) + C (x + 3)(x + 1) + D(x + 3)(x + 2) will be the numerator.

Can you go from here?
I got that bit, and i assume you get a simultaneous equation but i don't understand how you get two separate equations when there's only a 2 on the numerator
13. (Original post by AndyOC)
I got that bit, and i assume you get a simultaneous equation but i don't understand how you get two separate equations when there's only a 2 on the numerator
Put x = -1 then

2 = D(-1 + 3)(-1 + 2)

Then repeat for other values of x to get values of B and C
14. (Original post by Muttley79)
Put x = -1 then

2 = D(-1 + 3)(-1 + 2)

Then repeat for other values of x to get values of B and C
i see, didnt realise its that simple. thanks for the help
15. (Original post by AndyOC)
i see, didnt realise its that simple. thanks for the help
No problem

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