Question

Find A,B,C,a,b,c:
below is the equation

Answer 1

\[\frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c}\]

Answer 2

I tried expanding and simplifying but it was messier than I had thought.

Answer 3

The hint most probably in this question is that you do not actually have to solve it in the conventional sense.
You need to factorize the denominator in the LHS as
x(x-1)(x-2)
And then because ABCabc are constants you can just relate to the LHS and get the answer.
Understanding me? :)

Answer 4

So far I've got: [A(x-b)(x-c) + B(x-a)(x-c) +C(x-a)(x-b)]x(x-1)(x-2)=(x-a)(x-b)(x-c). Am I going down the wrong path?

Answer 5

I dont know how to execute it.

Answer 6

so it's likely I have gone down the wrong path.

Answer 7

Please give me another hint!

Answer 8

\(\large \frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \)

Answer 9

\(\large \frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \)

\(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \)

Answer 10

OK, will I just randomly asign 0,1,2 to a, b,c? and A+B+C=1 so will I have a system of equations solving for A, B, C?

Answer 11

exactly !!

Answer 12

A(x-1)(x-2) + Bx(x-2) + Cx(x-1) = 1

A+B+C = 1
A = 1/2
....

Answer 13

so is there a missing equation here? I only see 2 equations but am solving for A,B,C.

Answer 14

\(\large \frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \)

\(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \)

\(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A(x-1)(x-2) + Bx(x-2)+Cx(x-1) }{ x(x-1)(x-2) } \)

Answer 15

\(\large \frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \)

\(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \)

\(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A(x-1)(x-2) + Bx(x-2)+Cx(x-1) }{ x(x-1)(x-2) } \)

\(\large 1 = A(x-1)(x-2) + Bx(x-2)+Cx(x-1)\)

Answer 16

compare x^2, x coefficients and constant terms
u should get 3 equations

Answer 17

so one equation is A(x-1)(x-2) +Bx(x-2) + Cx(x-1) =1

and another is A + B + C = 1? I just don't see how to get to A = 1/2

Answer 18

here is the trick,

\(\large 1 = A(x-1)(x-2) + Bx(x-2)+Cx(x-1)\)

compare x^2 coefficients both sides
wat do u get ?

Answer 19

left side, x^2 coefficient = 0
right side, x^2 coefficient = ?

Answer 20

if its not clear, simply expand right side, and write it as polynomial

Answer 21

A(x^2 -3x +2) + B(x^2 -2x) + C(x^2 - x)

Answer 22

Yes, expand completely

Answer 23

\(\large 1 = A(x-1)(x-2) + Bx(x-2)+Cx(x-1)\)

\(\large 1 = A(x^2-3x+2) + B(x^2-2x)+C(x^2-x)\)

\(\large 1 = Ax^2-3Ax+2A + Bx^2-2Bx+Cx^2-Cx\)

\(\large 1 = x^2(A+B+C) + x(-3A-2B-C) + 2A\)

Answer 24

see if that looks right

Answer 25

Yes it does.

Answer 26

now simply compare x^2, x and constant coefficients both sides

Answer 27

compare \(x^2\) coefficients
0 = A+B+C ----------------------(1)


compoare \(x\) coefficients
0 = -3A-2B-C ----------------(2)

compare constant terms
1 = 2A --------------(3)

Answer 28

3 equations, 3 unknowns, u can solve :)

Enjoy..

Answer 29

I see thank you so much.

Answer 30

np :)