Question

please help me to solve this question,
Express as partial fraction (Repeated linear factors) : 1 / x^3(1-2x)

Answer 1

Any ideas?

Answer 2

For terms with a power higher than 1, like \(x^2\) for example, you could do this: \(\Large \frac{1}{x^2}=\frac{A}{x}+\frac{B}{x^2}\) ..

Answer 3

after get equation 1= A(x^2)(1-2x) + B(x)(1-2x) + C(1-2x) + D(x^3), how to find the value of A,B,C,D ? I try to substitute x as 0 , but cant find the ans..

Answer 4

I would approach this problem this way :
\(\Large \dfrac{1}{x^3 (1-2x)} =\dfrac{-1}{2}\dfrac{1}{x^3 (x-1/2)} \)

now for

\(\dfrac{1}{x^3 (x-1/2)}\)

we can decompose it like
\(\Large \dfrac{1}{x^3 (x-1/2)} = \dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x^3}+\dfrac{D}{x-1/2}\)

now make the common denominator and try to find A,B,C,D values
et me know if you get stuck...

Answer 5

yeah,then i get equation 1 = A1= A(x^2)(1-2x) + B(x)(1-2x) + C(1-2x) + D(x^3),
i stuck in finding the value of A,B,C,D.. is it need to expand the equation first? how?

Answer 6

ok, so you want to stick with 1-2x

anyways, you can find D and C easily

plug in x= 0 in the equation and you will get C
plug in x= 1/2 in the equation and you will get D
try :)

Answer 7

for A and B, yes, you need to expand by distributing.
then compare the co-efficients....if you don't know how, first find C,D which are easy to get.

Answer 8

i see.. so the value of C=1, D=8?

Answer 9

those are correct :)

Answer 10

then substitute value of C, D into the equation?

Answer 11

that would simplify things, yes.
and start distributing in the same step

Answer 12

\(1= A(x^2)(1-2x) + B(x)(1-2x) + C(1-2x) + D(x^3) \\ 1=Ax^2 -2Ax^3 +Bx-2Bx+C-2Cx+Dx^3\)

\(1=Ax^2 -2Ax^3 +Bx-2Bx+1-2x+8x^3\)

\(0=Ax^2 -2Ax^3 +Bx-2Bx-2x+8x^3\)

Answer 13

typing error :
\(1=Ax^2−2Ax^3+Bx−2Bx^2+1−2x+8x^3\)

Answer 14

do you get what i did there ?

Answer 15

aaa i see..yes i get it, then, to find A just subs x as 0?

Answer 16

you already plugged in x=0 to get C, now if you again plug in x=0 you get 0=0 :P

so now tell me the co-efficient of \(\Large x^3 \) on right side ??

Answer 17

ohh i see hehe. erm 2B?

Answer 18

ehh 2A

Answer 19

no...what is before x^3 ??

example in,
\(4Ay^2 +2y+By^2\)
the co-efficient of y^2 is 4A+B

so here, only 2A ?? or someting else too ?

Answer 20

8

Answer 21

sry i hv eye problem, -2A , -2B, 8

Answer 22

not 2B :P

Answer 23

its ok

not -2B, its with x^2, not x^3

so, co-efficient of x^3 on right side is
\(-2A +8\)
agree??
and on left its 0

so
-2A+8=0
you'll get A from here

Answer 24

then compare the co-efficients of x^2 on both sides, and you will get B
ask if any more doubts then.

Answer 25

erm why at the left side is 0, not 1? bit blur ;P

Answer 26

\(1=Ax^2 -2Ax^3 +Bx-2Bx^2+1-2x+8x^3\)
Subtracting 1 from both sides

\(0=Ax^2 -2Ax^3 +Bx-2Bx^2-2x+8x^3\)

got this step ?

Answer 27

and
\(0 = 0+0x+0x^2 +0x^3+....\)

Answer 28

so
\(0x^3 = -2Ax^3 +8x^3 = (-2A+8)x^3\)

so -2A+8= 0
hope this makes everything clear :)

Answer 29

ohhh now i clearly see.. like it.. ;P

Answer 30

similarly compare co-efficient of x^2 from both sides, and u get B too

Answer 31

so A=1/4, B=1/8?

Answer 32

how did u get A =1/4 ? :O

Answer 33

ohh, 4?

Answer 34

yes, now correct :)

Answer 35

and B=2?

Answer 36

\(\huge \checkmark \)

Answer 37

thank you very much helping me :)

Answer 38

welcome ^_^
happy to help :)

Answer 39

you seem to be new here, so
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 40

yes i am, thank you ^^