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Partial fractions c3 help..EASY!!

how do i solve this:

A(1-x) + B(2+x) + c(1-x)^2 = 1

I know to get B we let x=1 and solve to get B=1/3

But to get A and C do we have to equate coefficients?

Would equating x terms give: -A + B - 2C = 0 ?
and then equating constant terms: A + 2B + C = 1

then we add the two to remove the x term then sub 1/3 in for B to allow us to get C.

But I get that C=0?

How can this be? What have i done wrong?

(edited 10 years ago)

A(1 - x) + B(2 + x) + c(1 - x)^2 = 1

Let

x = 1: 3B = 1

B = \frac{1}{3}

Comparing coefficients of

x^2 : cx^2 = 0x^2

\therefore c = 0

Let

x = -2: 3A + 9C = 1

A = \frac{1 - 9C}{3}

A = \frac{1}{3}

A = \frac{1}{3}, B = \frac{1}{3}, C = 0

This is a slightly different method, but it gives

c = 0

again.

What was the original fraction?

OP

Original post by GingerCodeMan

A(1 - x) + B(2 + x) + c(1 - x)^2 = 1

Let

x = 1: 3B = 1

B = \frac{1}{3}

Comparing coefficients of

x^2 : cx^2 = 0x^2

\therefore c = 0

Let

x = -2: 3A + 9C = 1

A = \frac{1 - 9C}{3}

A = \frac{1}{3}

A = \frac{1}{3}, B = \frac{1}{3}, C = 0

This is a slightly different method, but it gives

c = 0

again.

What was the original fraction?

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

many thanks

You should have written

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

.

Tut-tut. Now C wont be zero.

(edited 10 years ago)

Understand?

{A \over 1-x} \times (1-x)^2 (2+x) = A (1-x) (2+x)

....But you wrote down

A (1-x)

instead. That's where the mistake started.

(edited 10 years ago)

Starting fom the beginning

{1 \over (1-x)^2(2+x)} = {A \over 1-x} + {B \over (1-x)^2} + {C \over 2+x}

Multiplying both sides by

(1-x)^2(2+x)

gives

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

.

Setting

x=1

gives

1=3B

. Setting

x=-2

gives

1=9C

. Lastly, equating

x^2

terms gives

0 = -A +C

.

So

A = {1 \over 9}

,

B = {1 \over 3}

and

C = {1 \over 9}

.

Original post by Mr Tall

how do i solve this:

A(1-x) + B(2+x) + c(1-x)^2 = 1

Original post by GingerCodeMan

A(1 - x) + B(2 + x) + c(1 - x)^2 = 1

Original post by steve44

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

.

\equiv

is how you get the identity sign

GingerCodeMan asked:

"What was the original fraction?"

Mr Tall said:

"

{1 \over (1-x)^2(2+x)}

"

This fraction leads to the equation

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

But Mr Tall had written

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

instead (look at the first term on the LHS of these two equations). Mr Tall made a typo here, this is the mistake!!!

You cant possibly end up with

C=0

(

C

corresponding to the term

{C \over (2+x)}

in the partial fraction expansion) because otherwise you would have

{1 \over (1-x)^2(2+x)}

is divergent for

x=-2

while

{A \over 1-x} + {B \over (1-x)^2}

wouldn't be!!

This is all about a typo - that's all I'm saying. Correct this typo and then you dont get C=0.

(edited 10 years ago)

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