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Is there a way to tell how to factor the denom. when doing a partial fraction?

Using z-transforms to solve difference equations.Is there an easy way to factor polynomials with two variables?Factor Completely.Understanding Why Partial Fractions WorksHow to use partial fractions with a cubic factor on bottom?The numerators in partial fraction decomposition involving repeated factorsPartial fraction decomposition of $frac1x^4-x^2$Do we have to memorize partial fraction decompositions?What is the form of partial fraction decomposition when the exponent is inside the factor?Cannot Understand how to Simplify Algebraic Fractions with Opposite Negatives

1

1

$begingroup$

My question comes from this specific problem from my homework:

$$fracx81x^4 - 1$$

Initially, I factored the denominator out to $(9x^2+1)(9x^2-1)$ and used this to find the $A,B,C,D$ to decompose the fraction, and everything seemed to be working fine but when I entered the final answer into the homework website it said it was wrong. The website had a video showing how to get the answer and they factored the denom. out into the $3$ factors instead of $2$.

This is probably the only thing that annoys me about partial fractions. Sometimes I can't tell if I should factor the denom. into $2$, $3$, or $4$ factors. If there is some sign I am missing here please do tell!

(Also, I do know that usually, you should factor it so there are as many factors as there are terms in the numerator. Or is that wrong? )

Thanks,

factoring partial-fractions

share|cite|improve this question

edited Mar 17 at 9:11

José Carlos Santos

170k23132238

asked Mar 6 at 23:11

Andrew LeonardAndrew Leonard

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Please use MathJax. Consider factoring $9x^2-1$
  $endgroup$
  – J. W. Tanner
  Mar 6 at 23:19
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Web-based assignment software is often useless, or worse: you might well have got a correct answer, just not the correct answer that it was expecting, and unlike a human, there's no leeway for alternative but equally correct answers. Presumably, the expected answer was the one in which the denominator had been factorised as far as it could go: $81x^4 - 1 = (9x^2+1)(3x-1)(3x+1)$ (notice you can't factorise any of these brackets further). But that doesn't mean your answer was wrong necessarily. It depends on the question and the context.
  $endgroup$
  – Billy
  Mar 6 at 23:20
  

  
  
  
  

add a comment | 

1

1

$begingroup$

My question comes from this specific problem from my homework:

$$fracx81x^4 - 1$$

Initially, I factored the denominator out to $(9x^2+1)(9x^2-1)$ and used this to find the $A,B,C,D$ to decompose the fraction, and everything seemed to be working fine but when I entered the final answer into the homework website it said it was wrong. The website had a video showing how to get the answer and they factored the denom. out into the $3$ factors instead of $2$.

This is probably the only thing that annoys me about partial fractions. Sometimes I can't tell if I should factor the denom. into $2$, $3$, or $4$ factors. If there is some sign I am missing here please do tell!

(Also, I do know that usually, you should factor it so there are as many factors as there are terms in the numerator. Or is that wrong? )

Thanks,

factoring partial-fractions

share|cite|improve this question

edited Mar 17 at 9:11

José Carlos Santos

170k23132238

asked Mar 6 at 23:11

Andrew LeonardAndrew Leonard

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Please use MathJax. Consider factoring $9x^2-1$
  $endgroup$
  – J. W. Tanner
  Mar 6 at 23:19
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Web-based assignment software is often useless, or worse: you might well have got a correct answer, just not the correct answer that it was expecting, and unlike a human, there's no leeway for alternative but equally correct answers. Presumably, the expected answer was the one in which the denominator had been factorised as far as it could go: $81x^4 - 1 = (9x^2+1)(3x-1)(3x+1)$ (notice you can't factorise any of these brackets further). But that doesn't mean your answer was wrong necessarily. It depends on the question and the context.
  $endgroup$
  – Billy
  Mar 6 at 23:20
  

  
  
  
  

add a comment | 

$begingroup$

My question comes from this specific problem from my homework:

$$fracx81x^4 - 1$$

Initially, I factored the denominator out to $(9x^2+1)(9x^2-1)$ and used this to find the $A,B,C,D$ to decompose the fraction, and everything seemed to be working fine but when I entered the final answer into the homework website it said it was wrong. The website had a video showing how to get the answer and they factored the denom. out into the $3$ factors instead of $2$.

This is probably the only thing that annoys me about partial fractions. Sometimes I can't tell if I should factor the denom. into $2$, $3$, or $4$ factors. If there is some sign I am missing here please do tell!

(Also, I do know that usually, you should factor it so there are as many factors as there are terms in the numerator. Or is that wrong? )

Thanks,

factoring partial-fractions

share|cite|improve this question

edited Mar 17 at 9:11

José Carlos Santos

170k23132238

asked Mar 6 at 23:11

Andrew LeonardAndrew Leonard

62

$endgroup$

share|cite|improve this question

edited Mar 17 at 9:11

José Carlos Santos

170k23132238

asked Mar 6 at 23:11

Andrew LeonardAndrew Leonard

62

share|cite|improve this question

edited Mar 17 at 9:11

José Carlos Santos

170k23132238

asked Mar 6 at 23:11

Andrew LeonardAndrew Leonard

62

edited Mar 17 at 9:11

José Carlos Santos

170k23132238

edited Mar 17 at 9:11

José Carlos Santos

170k23132238

asked Mar 6 at 23:11

Andrew LeonardAndrew Leonard

62

asked Mar 6 at 23:11

Andrew LeonardAndrew Leonard

62

1 Answer
1

active

oldest

votes

4

$begingroup$

The number of factors is not important. What is important is to decompose the denominator into irreducible factors. And, if you are working over $mathbb R$ or over $mathbb Q$, then, although $9x^2+1$ is irreducible, $9x^2-1$ is not, since it is equal to $(3x-1)(3x+1)$.

And if you were working over $mathbb C$, then you would have to write $9x^2+1$ as $(3x-i)(3x+i)$ too.

share|cite|improve this answer

edited Mar 6 at 23:39

J. W. Tanner

3,8871320

answered Mar 6 at 23:20

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

add a comment | 

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4

$begingroup$

The number of factors is not important. What is important is to decompose the denominator into irreducible factors. And, if you are working over $mathbb R$ or over $mathbb Q$, then, although $9x^2+1$ is irreducible, $9x^2-1$ is not, since it is equal to $(3x-1)(3x+1)$.

And if you were working over $mathbb C$, then you would have to write $9x^2+1$ as $(3x-i)(3x+i)$ too.

share|cite|improve this answer

edited Mar 6 at 23:39

J. W. Tanner

3,8871320

answered Mar 6 at 23:20

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

add a comment | 

4

$begingroup$

The number of factors is not important. What is important is to decompose the denominator into irreducible factors. And, if you are working over $mathbb R$ or over $mathbb Q$, then, although $9x^2+1$ is irreducible, $9x^2-1$ is not, since it is equal to $(3x-1)(3x+1)$.

And if you were working over $mathbb C$, then you would have to write $9x^2+1$ as $(3x-i)(3x+i)$ too.

share|cite|improve this answer

edited Mar 6 at 23:39

J. W. Tanner

3,8871320

answered Mar 6 at 23:20

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

add a comment | 

$begingroup$

The number of factors is not important. What is important is to decompose the denominator into irreducible factors. And, if you are working over $mathbb R$ or over $mathbb Q$, then, although $9x^2+1$ is irreducible, $9x^2-1$ is not, since it is equal to $(3x-1)(3x+1)$.

And if you were working over $mathbb C$, then you would have to write $9x^2+1$ as $(3x-i)(3x+i)$ too.

share|cite|improve this answer

edited Mar 6 at 23:39

J. W. Tanner

3,8871320

answered Mar 6 at 23:20

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

share|cite|improve this answer

edited Mar 6 at 23:39

J. W. Tanner

3,8871320

answered Mar 6 at 23:20

José Carlos SantosJosé Carlos Santos

170k23132238

edited Mar 6 at 23:39

J. W. Tanner

3,8871320

edited Mar 6 at 23:39

J. W. Tanner

3,8871320

answered Mar 6 at 23:20

José Carlos SantosJosé Carlos Santos

170k23132238

answered Mar 6 at 23:20

José Carlos SantosJosé Carlos Santos

170k23132238