Question

Show that \[({2 \over (x+1)(x+3)})^2 = {1 \over (x+1)^2} - {1 \over x+1}+ { 1 \over x+3} + {1 \over (x+3)^2}\]

Answer 1

You will need 4 variables.

A/(x+1) + B/(x+1)^2 + C/(x+3) + D/(x+3)^

and the usual process ... and plus it's going to be ugly, best of luck

Answer 2

*A/(x+1) + B/(x+1)^2 + C/(x+3) + D/(x+3)^2

Answer 3

Could you help me through this ugly process?

Answer 4

Please :D

Answer 5

it's going to be MUCH easier if you solve it on paper... here's quite difficult ! ... you're stuck or you're lazy ? :P

Answer 6

stuck. not lazy :P

Answer 7

... Ok, give me a minute :F

Answer 8

I'll try (if I can) lol

Answer 9

:D

Answer 10

yes ... it's quite going to be ugly, ... since it's going be lots of simplification ... @Kreshnik will you help her???

Answer 11

...I'll try, but as you know , my math sucks LOL

Answer 12

Thanks for trying :D Your math doesn't suck :p

Answer 13

keep trying ... this is a 10 or 11 grade question ... it should be pretty easy as you are doing other advanced stuff.

Answer 14

Haha, it's actually a 12th grade question :/

Answer 15

anyway keep trying ...it;s just simplification.

Answer 16

lol, I know, Let me go grab a camera, I'll upload a photo :D

Answer 17

can you continue? ... (It's not hard,) I have to go, Be right back.

Answer 18

How did you get the numbers at the top? Wouldn't it be A/... +B/.... +C/.... + D/...?

Answer 19

@Kreshnik thats not right.

Answer 20

experiment, how is it done?

Answer 21

yes, you have to simplify interns of A(x+1)(x+3)^2 +B(x+3)^2 + C(x+1)^2(x+3)+D(x+1)^2

simplify this .... you should practice these kinda of stuff.

Answer 22

Would you first open the brackets like this?\[({2 \over (x+1)(x+3)})^2---> {4 \over (x+1)(x+3)(x+1)(x+3)}?\]

Answer 23

ahhhh :@ .. I multiplied, I'll do it again. hold on

Answer 24

would it be\[{4 \over (x+1)(x+3)(x+1)(x+3)}= {A \over x+1}+{B \over x+3}+{C \over x+1} + {D \over x+3}?\]

Answer 25

@experimentX

Answer 26

put square below on B, and C

Answer 27

How come?

Answer 28

\[{4 \over (x+1)(x+3)^2(x+1)^2(x+1)}={A \over x+1} +{B \over (x+3)^2}+{C \over (x+1)^2}+ {D \over c+3}?\]

Answer 29

The last = x+3 sorry*

Answer 30

Would that be the start? @experimentX

Answer 31

yes,, best of luck

Answer 32

well, unfortunately, the math I'm doing, seems to be wrong :( .. don't we just have to solve for x? :(:(

Answer 33

well, no ... it's collecting coefficients of x's

Answer 34

but what happens if we solve for x? ... I got 0=0 !!

Answer 35

\[4=A (x+3)^2(x+1)^2(x+3) + B(x+1)(x+1)^2(x+3)+C(x+1)(x+3)^2(x+3)\]\[+D(x+1)(x+3)^2(x+1)^2\]

Answer 36

?

Answer 37

Ah ... no you overdid it.

Answer 38

So.. How?

Answer 39

A(x+1)(x+3)^2 +B(x+3)^2 + C(x+1)^2(x+3)+D(x+1)^2

Jeez ... what do kids learn these days at school.

Answer 40

I'm sorry to take your time, obviously I'm 100% dumb, I wish I could help. :( I'm out .

Answer 41

Umm. how did you get that?

Answer 42

@Kreshnik you can give a try ...

Answer 43

I didn't learn it at school :o...

Answer 44

Taking LCM

Answer 45

@experimentX I don't know these kind of tasks, I wish I did. Why should I try, when I'm always wrong :(

Answer 46

Is there a common lcm for this then?

Answer 47

Can you explain more how you got that?

Answer 48

yes (x+1)^2(x+3)^2