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# I have proved mathsnet.net wrong! (partial fractions) watch

1. Attached in the attachment is their solution, and here is mine. Tell me if i really have, or whether i am being a complete idiot (usually the case)

The question is :

5x³ + 8
x(x-2)

Firstly i divided it as it is an improper fraction, and i got (x² - 2x)(5x+10) remainder 10x +8

So, A(x-2) + B(x) = 10x+8
x=2
2B = 28
B=14

if x = 0
-2A= 8
A=-4

Therefore, 5x +10 -4/x + 14/(x-2)

Thanks for looking over it.
Attached Images

2. Your solution is incorrect, theirs is correct.
3. (Original post by Library)
Attached in the attachment is their solution, and here is mine. Tell me if i really have, or whether i am being a complete idiot (usually the case)

The question is :

5x³ + 8
x(x-2)

Firstly i divided it as it is an improper fraction, and i got (x² - 2x)(5x+10) remainder 10x +8

So, A(x-2) + B(x) = 10x+8
x=2
2B = 28
B=14

if x = 0
-2A= 8
A=-4

Therefore, 5x +10 -4/x + 14/(x-2)

Thanks for looking over it.

your remainder should have been 20x + 8
4. *Suprised* why?... what have i done wrong?
5. you can check to see if it is working by putting a value for x into both versions...eg if x=3 then the first version is worth 143/3 but the second version is worth 113/3 so there must be an error somewhere.

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Updated: November 15, 2005
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