Question

Express Partial fraction
x^2+x-1/(x+1)(x+2)

Answer 1

@Hero

Answer 2

familiar with long division ?

Answer 3

yes

Answer 4

p(x) / q(x)

degree of p(x) needs to be less than q(x) for partial fractions to work

Answer 5

but in your rational function, both degrees are equal
divide so that the degree of numerator becomes less than the denominator

Answer 6

p(x) is the numerator

Answer 7

I know this method

Answer 8

Oh, oh! Is this partial fraction decomposition? :D

Answer 9

Yes

Answer 10

Okay, what you need to do is just completely ignore the top for now, and simplify the bottom as much as possible.

So it's \(\Large \frac{x^2+x-1}{(x+1)(x+2)}\) right?

Answer 11

ys

Answer 12

Alright, so now just break \(\Large \frac{x^2+x-1}{(x+1)(x+2)}\) into fractions with A and B and C's, etc, just as many as you need to match the denominator with a term:

\[\Large \frac{x^2+x-1}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}\]

^That's how you would set it up as. Just a rule, whatever degree power you have at the bottom, the top will be one less of that. SO Since the bottom had a degree of 1 for x, the top would of been a power of 0, which is just 1, so A*1=A.

Answer 13

If you wanna solve it, just multiply it out by the common denominators

Answer 14

...=A(x+2)+B(x+1)

Answer 15

Exactly!

\(\Large x^2+x-1=A(x+2)+B(x+1)\)

Use the left equation as a guide for values if you're gonna distribute out the right :)

Answer 16

when x= -2
(-2)^2+(-2)-1=A(0)-B
B=-1
when x=-1
A=1

Answer 17

So,

Answer 18

1/(x+1)- 1/(x+2)

Answer 19

If you plug in -1 you'll get \(\Large (-1)^2-1-1=A\),

\(\Large (1-1)-1=A\)
\(\Large 0-1=A\)

So -1 for A too :P

Answer 20

okay

Answer 21

Hw wld the final answer be

Answer 22

What you just wrote with a -1 as A, or you could factor out that negative altogether

\(\Large -(\frac{1}{x+1}+\frac{1}{x+2})\)

Answer 23

Based on the answer from the book,
it's 1- 1/(x+1)- 1/(x+2)
So, is the book answer wrong ?

Answer 24

?

Answer 25

No, the book isn't wrong, we just forgot to simplify in the beginning.. notice that if we FOIL, they will both have the same x power.

Answer 26

So we actually needed to composite this:
\[\Large \frac{x^2}{x^2+3x+3}+\frac{x-1}{(x^2+3x+3)}\]

Which is \(\Large 1+(\frac{x-1}{x^2+3x+3}) \)

Answer 27

So we just needed the simplification, in the beginning, everything else is good :)

Answer 28

Okay, understood!
Thanks Luigi

Answer 29

You're welcome, good luck bud~

Answer 30

\(\Large \frac{x^2+x-1}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}\)

that is wrong^^
as rational said,
partial fractions do not work when numerator degree is EQUAL or HIGHER than denominator's

Answer 31

Yea I noticed that, we forgot the simplification in the beginning :l