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Partial fractions maths!

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Reply 40
Avatar for Jane122
Jane122
OP
2
Original post by RDKGames
Ax+B+Cx+1Ax+B+\frac{C}{x+1}


Can you look@ my third attachment
Reply 41
Avatar for h3rmit
h3rmit
15
Original post by RDKGames
Ax+B+Cx+1Ax+B+\frac{C}{x+1}

How do you figure out which general partial fraction form you need for a fraction?
Original post by Jane122
Im a female. So shut up.


Aren't you a feisty one.

I literally just told you to simplify it and you will get 4A+B=12 and/or use x=1, 2
Reply 43
Avatar for Jane122
Jane122
OP
2
Original post by Hamzah249
Aren't you a feisty one.

I literally just told you to simplify it and you will get 4A+B=12 and/or use x=1, 2



Dont assume then. Jane is not a male name, it's an idiotic assumption. I've already got the correct answer. Thanks.
Reply 44
Avatar for Jane122
Jane122
OP
2
IMG_4894.jpg

With part B, I have this so far


Help @Hamzah249
Original post by h3rmit
How do you figure out which general partial fraction form you need for a fraction?


That's what I was actually asking
Original post by Jane122
Dont assume then. Jane is not a male name, it's an idiotic assumption. I've already got the correct answer. Thanks.


Why are you so frustrated?
Reply 47
Avatar for Jane122
Jane122
OP
2
Original post by student1256
why are you so frustrated?


because nobody is looking at my attachment and im stuck
Original post by h3rmit
How do you figure out which general partial fraction form you need for a fraction?


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)+Qp(x)gm(x)\displaystyle \frac{f_n(x)}{g_m(x)}=R_{n-m}(x)+\frac{Q_{p}(x)}{g_m(x)} for n≥mn\geq m where R,QR,Q are polynomial functions of their respective degrees and p<mp<m.
(edited 7 years ago)
Original post by Jane122
because nobody is looking at my attachment and im stuck

Are you still stuck? I thought you understand now with rdk's help
Original post by Muttley79
Please don't post full solutions - the OP would have got there :smile:


No he wouldn't have :wink:
Original post by Jane122
IMG_4894.jpg

With part B, I have this so far


Help @Hamzah249


You call him idiotic then go on to tag him asking for help, really?... lmao
Reply 52
Avatar for Jane122
Jane122
OP
2
Original post by Student1256
Are you still stuck? I thought you understand now with rdk's help


I don't I understand what he wrote.

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.


I'm confused
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.


Now that's sexy :biggrin:
Original post by Jane122
IMG_4894.jpg

With part B, I have this so far


Help @Hamzah249


You've shown in part a that -3 is a root, meaning that (x-3) is a factor of f(x)=0.
Divide f(x) by this using polynomial division (c2 stuff). Then you should have a nice quadratic to factorise. Following that it should be a simple partial fraction question without any of the nasty comparing coefficients needed.
Edit: x+3 not x-3 pardon.
(edited 7 years ago)
RDK already helped you with, gurl, you want 5 people to tell you same thing to get it through your head.
Original post by Jane122


I'm confused


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.
Reply 57
Avatar for Jane122
Jane122
OP
2
Original post by Student1256
You call him idiotic then go on to tag him asking for help, really?... lmao


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
Original post by Jane122
IMG_4894.jpg

With part B, I have this so far


Help @Hamzah249


Can you show your working please?
Reply 59
Avatar for Jane122
Jane122
OP
2
IMG_4897.jpg
Original post by Mr M
Can you show your working please?


Finally someone intelligent !

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