Question

Partial Fractions Decomposition

Answer 1

\[\frac{ x ^{2}+2x+3 }{ (x+1)(x ^{2}+2x+4) }\]

\[x ^{2}+2x+3 = Ax ^{2}+Bx ^{2}+2ax+4a+bx+cx+c\]

Answer 2

So would it be

a+b =1
2x+4=1
b+c=1
c=0
@agent0smith

Answer 3

\(x ^{2}+2x+3 = Ax ^{2}+Bx ^{2}+2ax+4a+bx+cx+c\)
\((1)x ^{2}+(2)x+(3) = (Ax ^{2}+Bx ^{2})+(2ax+bx+cx)+(4a+c)\)
\((1)x ^{2}+(2)x+(3) = (A+B)x ^{2}+(2a+b+c)x+(4a+c)\)
1=A+B
2=2a+b+c
3=4a+c

Answer 4

Thanks @Zale101

Answer 5

No problem!

Answer 6

1=A+B
2=2a+b+c
3=4a+c

Solve the top equation for B in terms of A
Solve the bottom for C in terms of A.
Plug into the middle equation to find A.

Answer 7

I mean i multiplied the top by -2

Answer 8

0=-b+c so now we plug it in to the 3rd one

Answer 9

Why not do what I said in the last post...

Answer 10

1. top equals B=-A
what do you mean by 2

Answer 11

No it doesn't.

1=A+B <=== solve for B
2=2a+b+c
3=4a+c <=== solve for c

Answer 12

i forgot the plus 1 lol

Answer 13

i got 2/3

Answer 14

so b would be 1/3

Answer 15

Now you can find the others, first A then C.

Answer 16

A=2/3
b=1/3
c=1/3

Answer 17

https://www.wolframalpha.com/input/?i=1%3DA%2BB,+2%3D2A%2BB%2BC,+3%3D4A%2BC

Answer 18

This is a reason why you dont take chem and calc together for 6weeks lol


Thanks for the help @agent0smith cleared it up for me :D

Answer 19

I think it might have been faster to use the way i suggested last night, plugging in numbers for x should always be the first priority

Answer 20

But that doesn't always work though

Answer 21

But always look for it first.\[\large \frac{ x ^{2}+2x+3 }{ (x+1)(x ^{2}+2x+4) } = \frac{ A }{x+1 }+\frac{ Bx+C }{x^2+2x+4 }\]
\[\large x^{2}+2x+3= A(x^2+2x+4) + (Bx+C)(x+1)\]
Plugging in x=-1 immediately yields A.

Giving a system of two equations instead of 3.

Answer 22

Then plugging in x=0 gets you C...

Answer 23

Then just plug in any random value of x to find B. No systems of equations needed.

Answer 24

do you think 0 would work

Answer 25

If you plug in x=-1 you can find A.

Then knowing A, if you plug in x=0, you can find C.

Then plug in any other value of x (not 0 or -1) to find B, I'd use x=1 for simplicity.

Answer 26

c =4a-3

Answer 27

Ill probably stick to the first one and use this one when its usable llol

Answer 28

Well you were supposed to find A before trying to find C :P

You just have to eyeball the equation and see if there's any x's you can plug in that eliminate any of the unknowns. It'll often save you lots of time... I used to hate partial fractions till I discovered this method.

Answer 29

Yea i learned this in highschool but forgot everything lol thanks for the tip