MedVision ad

Express as partial fraction (1 Viewer)

Nashchnc

New Member
Joined
Nov 7, 2013
Messages
16
Gender
Male
HSC
2014
I don't have an issue with any of it, but this one question is totally messing with my head, i get different answers by equating coefficients and by letting x=-1. Just totally messing with me :(


(4x2-2x)/(x+1)(x2+1)
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,384
Gender
Male
HSC
2006
I don't have an issue with any of it, but this one question is totally messing with my head, i get different answers by equating coefficients and by letting x=-1. Just totally messing with me :(


(4x2-2x)/(x+1)(x2+1)
Did you get it in this form?

 

Nashchnc

New Member
Joined
Nov 7, 2013
Messages
16
Gender
Male
HSC
2014
That would be wherein the problem lies... I dont quite understand why though?
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,384
Gender
Male
HSC
2006
That would be wherein the problem lies... I dont quite understand why though?
Partial fractions should be expressed where the degree of the numerator is always less than the degree of the denominator. It is effectively a polynomial division. If the degree of the numerator is greater than or equal to the degree of the denominator then we can reduce it further by polynomial division until the degree of the remainder term falls below the degree of the divisor.

In the case of a quadratic factor (degree 2) which has no real roots (i.e. cannot be factorised further under real numbers), the numerator must be either degree 0 (i.e. a constant) or degree 1. Since we can't be sure exactly which degree the numerator is at first, we always set it to the highest degree (i.e. bx + c which is degree 1) for generalisation and should it actually be a degree zero then the working gives b = 0 for us.
 
Last edited:

hit patel

New Member
Joined
Mar 14, 2012
Messages
568
Gender
Male
HSC
2014
Uni Grad
2018
Trebla are you sure? i Got a= 3 b=1 and c=-3 which works out completely. WHat you do is use the substitution method to work out A and then use the Equation coefficients method to work out b and c by inspection?


Answer comes out to be 3ln(x+1) +(1/2)(ln(x^2+1)) -3 tan^-1 (x) + C
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,384
Gender
Male
HSC
2006
Trebla are you sure? i Got a= 3 b=1 and c=-3 which works out completely. WHat you do is use the substitution method to work out A and then use the Equation coefficients method to work out b and c by inspection?


Answer comes out to be 3ln(x+1) +(1/2)(ln(x^2+1)) -3 tan^-1 (x) + C
What do you mean? I just provided a general statement without actually doing the question.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top