Partial Fractions problem

Jul 2009
47
0
Singapore
Good Day,

I came across this partial fractions problem yesterday and I'm confused as to why I get the wrong answer although the factorising seems correct to me. I'll list down what I did and what is the answer provided by the book.

The steps I took to break it up
x / (1-x^2)(x-1)
= x / (1+x)(1-x)(x-1)
= -x / (x-1)^2 (1+x)
= - [1/4(x-1)] - [1/2(x-1)^2] + [1/4(1+x)]

However, the book states that the answer is [1/4(1-x)] - [1/2(1-x)^2] + [1/4(1+x)].

When I tested my answer and the book's, I found out that the book's answer gave the fraction that I was asked to break up.

Is there a reason why I must factorise the denominator into (1-x)^2 (1+x) and not (x-1)^2 (1+x)?

Thanks in advance.
 
Sep 2008
1,261
539
West Malaysia
Good Day,

I came across this partial fractions problem yesterday and I'm confused as to why I get the wrong answer although the factorising seems correct to me. I'll list down what I did and what is the answer provided by the book.

The steps I took to break it up
x / (1-x^2)(x-1)
= x / (1+x)(1-x)(x-1)
= -x / (x-1)^2 (1+x)
= - [1/4(x-1)] - [1/2(x-1)^2] + [1/4(1+x)]

However, the book states that the answer is [1/4(1-x)] - [1/2(1-x)^2] + [1/4(1+x)].

When I tested my answer and the book's, I found out that the book's answer gave the fraction that I was asked to break up.

Is there a reason why I must factorise the denominator into (1-x)^2 (1+x) and not (x-1)^2 (1+x)?

Thanks in advance.
hi

you are correct and the book is also correct. You just play with the algebra here.

-1/4(x-1)= -1/4(-(1-x))=1/4(1-x)

It doesn't matter since both forms are correct.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
\(\displaystyle (1- x)^2= (-(x-1))^2= (-1)^2(x-1)^2= (x-1)^2\)
 
Last edited by a moderator:
Jul 2009
47
0
Singapore
Great! Thanks for the quick response, everyone!