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Partial fractions maths!

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Original post by RDKGames
For a general polynomial function fn(x)f_n(x) of order nn it is a theorem that fn(x)gm(x)=Rn−m(x)+Qm−1(x)gm(x)\displaystyle \frac{f_n(x)}{g_m(x)}=R_{n-m}(x)+\frac{Q_{m-1}(x)}{g_m(x)} for n≥mn\geq m where R,QR,Q are polynomial functions of their respective degrees.



One thing though. Doesn't that general equation not leave space for the B, which is a constant?
Original post by Student1256
Now that's sexy :biggrin:


Seriously I'm done with my A2 maths and even I don't get it.
Original post by RDKGames
That last one was not for you unless you fully understand the concept of partial fractions which doesn't seem like it at this point.

In regards to your attachment, you have found the left over quadratic when you divided the cubic by x+3x+3. Now factorise that quadratic then express your cubic in terms of 3 the linear factors you end up with.


If you give this guy rep I deserve some too. Same answer but further up. Just cos he can use TeX :frown:
Original post by Jane122
I didn't call him idiotic, I said his assumption was idiotic. There's a difference , if you weren't too busy flirting with him and RDK


Hahahahahahahahaha this is some good ass trolling my friend
Original post by Jane122
IMG_4897.jpg

Finally someone intelligent !


Right that's wrong so we need to go back a step.

Can you factorise f(x) given the information provided in part a)?
Original post by Student1256
One thing though. Doesn't that general equation not leave space for the B, which is a constant?


It does, the constant is part of the function R1(x)R_1(x) when referring to the theorem.

Remember, a general polynomial is of form f(x)=anxn+an−1xn−1+an−2xn−2+...+a1x+a0f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0 for ai∈Qa_i\in \mathbb{Q} where a0a_0 is a constant so it is included.
Original post by Jane122
IMG_4887.jpg


Original post by Jane122
I swear nobody on this thread bothered to look AT THE PIC.


If you notice your pic is sideways, which isn't necessarily your fault (TSR Is annoying about it) but there is a correlation between that and your post, I suppose.

One thing you can do to guarantee correct orientation is to upload the image to Imgur and then copy the location and use img tags, or if that fails (it is a bit too long winded to be honest unless you have done it before) is to take multiple by changing the orientation of the phone/camera device and then find one that makes them right way up.
Original post by Hamzah249
Seriously I'm done with my A2 maths and even I don't get it.


Yeah it took me a while to get it but that's how they explain sh*t in undergraduate level math at some unis. It's quite a pain to understand sometimes but you get used to the format and wording after a while hamxa bhai. I'm in a level
Original post by Student1256
One thing though. Doesn't that general equation not leave space for the B, which is a constant?


I was thinking the same thing.
And looking at examples it's kinda hit and miss whether the constant is included. (I understand it's only in (x+n)^a)
Original post by SeanFM
If you notice your pic is sideways


I have a meme for that.
Rotate.jpg18.8KB
Original post by Student1256
Yeah it took me a while to get it but that's how they explain sh*t in undergraduate level math at some unis. It's quite a pain to understand sometimes but you get used to the format and wording after a while hamxa bhai. I'm in a level


Yeah I got it too after a while, just from a stand point of someone who can't get through partial fractions, it's basically impossible to get what he wrote.

Oh are you taking any modules in this month?
Original post by RDKGames
It does, the constant is part of the function R1(x)R_1(x) when referring to the theorem.

Remember, a general polynomial is of form f(x)=anxn+an−1xn−1+an−2xn−2+...+a1x+a0f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0 for ai∈Qa_i\in \mathbb{Q} where a0a_0 is a constant so it is included.


Ok so am i right in saying that this formula covers for the existence of a constant. But doesn't have any way of affirming when it is 0 (ie no constant). So in a question when we aren't given a format for the answer (like a/(x+g) + b/(x+h) + c), assume the existence of a constant?

A partial fractions question, that is.
(edited 7 years ago)
Original post by RDKGames
It does, the constant is part of the function R1(x)R_1(x) when referring to the theorem.

Remember, a general polynomial is of form f(x)=anxn+an−1xn−1+an−2xn−2+...+a1x+a0f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0 for ai∈Qa_i\in \mathbb{Q} where a0a_0 is a constant so it is included.


I see! Pure mathematics is beautiful and good job with your work on TSR
Original post by Hamzah249
Yeah I got it too after a while, just from a stand point of someone who can't get through partial fractions, it's basically impossible to get what he wrote.

Oh are you taking any modules in this month?


Hahahahahaha yeah for someone who is finding it difficult comparing coefficients (no offence to that person, I myself was pretty bad at maths) trying to understand that is like trying to decipher ancient hieroglyphs :laugh:
Original post by carpetguy
Ok so am i right in saying that this formula covers for the existence of a constant. But doesn't have any way of affirming when it is 0 (ie no constant). So in a question when we aren't given a format for the answer (like a/(x+g) + b/(x+h) + c), assume the existence of a constant?

A partial fractions question, that is.


Yes it covers for the existence of a constant that is 0 or not.

If you're not provided with a format, assume that there is a constant then you will eventually show through working out that it is either identically 0 or some other value.
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Original post by Mr M
Right that's wrong so we need to go back a step.

Can you factorise f(x) given the information provided in part a)?



I got 2x^2+3x+1 when I factorised it
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Original post by Student1256
Hahahahahahahahaha this is some good ass trolling my friend


What are you on about ?
Original post by Jane122
I



I got 2x^2+3x+1 when I factorised it


Right you need to factorise this quadratic too.

You'll then get:

10(2x+1)(x+1)(x+3)=A2x+1+Bx+1+Cx+3\displaystyle \frac{10}{(2x+1)(x+1)(x+3)} = \frac{A}{2x+1} + \frac{B}{x+1} + \frac{C}{x+3}

Now find A, B and C using your preferred method.
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Original post by Mr M
Right you need to factorise this quadratic too.

You'll then get:

10(2x+1)(x+1)(x+3)=A2x+1+Bx+1+Cx+3\displaystyle \frac{10}{(2x+1)(x+1)(x+3)} = \frac{A}{2x+1} + \frac{B}{x+1} + \frac{C}{x+3}

Now find A, B and C using your preferred method.


But I thought you wouldn't write it in the form of 0/x+3 with a remainder of 2x^2+3x+1 and do partial fractions on on the 0/x+3 :frown:
Original post by Jane122
But I thought you wouldn't write it in the form of 0/x+3 with a remainder of 2x^2+3x+1 and do partial fractions on on the 0/x+3 :frown:


Nope. Try and get A, B and C but, if you really have no idea what is going on, you probably should ask your teacher to go through it with you again.

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