Partial Fractions: Why Does (x+1)2(2x+1) Need 3 Terms?

Click For Summary

Discussion Overview

The discussion centers on the reasoning behind the necessity of including three terms in the partial fraction decomposition of a rational expression with repeated linear factors in the denominator, specifically in the context of the expression (11x² + 14x + 5)/[(x+1)²(2x+1)]. Participants explore the implications of the structure of the denominator on the form of the numerators in the decomposition.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions why the decomposition must include terms A/(x+1), B/(x+1)², and C/(2x+1), expressing confusion over an alternative approach that omits the first term.
  • Another participant suggests that omitting 1/(x+1) would lead to an incorrect form for the numerator of 1/(x+1)², which should be of the form Ax + B.
  • Several participants emphasize that the coefficients in the numerators are not known until the decomposition is completed, highlighting the exploratory nature of the process.
  • There is a discussion about the nature of the denominator factors, with some participants noting that while (x+1) is linear, its multiplicity affects the form of the numerator in the decomposition.
  • One participant points out that if the denominator were a non-factorable quadratic, the numerator would take a different form, such as Bx + D, indicating a distinction in treatment based on the factorization of the denominator.
  • Another participant agrees with the necessity of including the term B/(x+1)², reinforcing the idea that the structure of the denominator dictates the form of the numerators.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of including specific terms in the partial fraction decomposition, with no consensus reached on the alternative approaches proposed. The discussion remains unresolved regarding the implications of omitting certain terms.

Contextual Notes

Some participants reference the need for proper fractions and the importance of finding the lowest common denominator, indicating that the discussion may be limited by assumptions about the form of the original rational expression.

Deep_Thinker97
Messages
11
Reaction score
0
Why, when a fraction has repeated linear terms in its denominator e.g. (11x2+14x+5)/[(x+1)2(2x+1)] does it have to be split into three partial fractions, A/(x+1) + B/(x+1)2 + C/(2x+1)?
When my first saw this example, my initial reaction was to split it into A/(x+1)2 +B/(2x+1), but after working through this, I realized my method was wrong. Why doesn't it work? I don't want a worked answer to the example because I already know what it is. I just want a genuine logical reason to why the former works and the latter doesn't.
Thanks :)
 
Physics news on Phys.org
If you leave out 1/(x+1) then the numerator for 1/(x+1)² will be of the form Ax+B.
 
  • Like
Likes   Reactions: symbolipoint
Deep_Thinker97,
Try making a couple of simple examples.
Partial fraction decomposition is a way to find what the original fractions were before they were added (or subtracted).

What must be done to sum 3/(x+1)+8/(x+1)^2 ?

How about sum of 3/(x+1)+(8x+2)/(x+1)^2 ?

Try sum of 3/(x+1)+0/(x+1)^2 ?You always start with proper fractions.
In summing the original fractions, you need to bring some of them to higher terms in order to sum, so you need to find the lowest common denominator.

When you want to decompose a rational expression (partial fraction decomposition), you do not know in advance what are the coefficients or constants in the numerators of the decomposed parts.
 
mathman said:
If you leave out 1/(x+1) then the numerator for 1/(x+1)² will be of the form Ax+B.
Yes. One does not know the values of the coefficient and constant term until one has finished the decomposition work.

Excuse my comment there, since what I said was not exactly what should have been.
 
Last edited:
mathman said:
If you leave out 1/(x+1) then the numerator for 1/(x+1)² will be of the form Ax+B.
Checking about that required some review or restudy. The denominator (x+1) is linear, but it is used as multiplicity of 2; still it is a LINEAR factor. The numerator used for the partial fraction will be just a single constant. If the denominator factor were QUADRATIC AND NOT FACTORABLE in real numbers, then that numerator would be of the form Ax+B.
 
symbolipoint said:
Checking about that required some review or restudy. The denominator (x+1) is linear, but it is used as multiplicity of 2; still it is a LINEAR factor. The numerator used for the partial fraction will be just a single constant. If the denominator factor were QUADRATIC AND NOT FACTORABLE in real numbers, then that numerator would be of the form Ax+B.
Nope - why not factorable?.
a/(x+1)+b/(x+1)^2=(ax+(a+b))/(x+1)^2
 
mathman said:
Nope - why not factorable?.
a/(x+1)+b/(x+1)^2=(ax+(a+b))/(x+1)^2
The denominator factor of x+1 is linear. The original rational expression to decompose would contain a denominator (x+1)(x+1)=(x+1)^2=x^2+2x+1, the right side being a FACTORABLE quadratic.

All I could try now is to take the original poster's rational expression and work it on paper to decompose into its partial fractions.
 
Last edited:
Deep_Thinker97 said:
A/(x+1) + B/(x+1)2 + C/(2x+1)?
I agree with the second partial fraction expression with the constant, B, because the denominator is from having the linear x+1 with exponent 2.

Multiplying by the denominator of the original expression gives
11x^2+14x+5=A(x+1)(2x+1)+B(2x+1)+c(x+1)^2

Working with the right hand side,
arranging in decreasing terms of x,
(2A+C)x^2+(3A+2B+2C)x+(A+B+C)My best understanding here is that if denominator had a non-factorable x^2+mx+n, then this needs a numerator like Bx+D for a partial fraction.

Deep_Thinker97 already found his answers.
 

Similar threads

MHB Partial fractions (5x^2+1)/[(3x+2)(x^2+3)]
  • · Replies 3 ·
Replies
3
Views
3K
MHB Solve Int Partial Fractions: $\int\frac{6{x}^{2}+22x-23} {(2x-1)(x+3)(x-2)} dx$
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Partial Fraction Decomposition With Quadratic Term
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Differentiability of a Multivariable function
  • · Replies 1 ·
Replies
1
Views
3K
MHB 242t.8.5.9 expand the quotient by partial fractions
  • · Replies 3 ·
Replies
3
Views
2K
MHB Solving Partial Fractions & Maclaurin Series Q&A
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Understanding Mixed Partial Derivatives: How Do You Solve Them?
  • · Replies 3 ·
Replies
3
Views
2K
MHB Integration by Partial Fraction Decomposition - Yahoo Answers
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Partial Fractions: Exponent on Denominator Explained
  • · Replies 10 ·
Replies
10
Views
3K
A doubt in Partial fraction decomposition
  • · Replies 8 ·
Replies
8
Views
2K