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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
$begingroup$
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
edited Mar 17 at 9:11
José Carlos Santos
170k23132238
asked Mar 6 at 23:11
Andrew LeonardAndrew Leonard
62
$endgroup$
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
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
edited Mar 17 at 9:11
José Carlos Santos
170k23132238
asked Mar 6 at 23:11
Andrew LeonardAndrew Leonard
62
$endgroup$
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
factoring partial-fractions
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
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
edited Mar 17 at 9:11
José Carlos Santos
170k23132238
170k23132238
asked Mar 6 at 23:11
Andrew LeonardAndrew Leonard
62
asked Mar 6 at 23:11
Andrew LeonardAndrew Leonard
62
asked Mar 6 at 23:11
Andrew LeonardAndrew Leonard
62
62
$begingroup$
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 |
$begingroup$
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
$begingroup$
Welcome to Math Stack Exchange. Please use MathJax. Consider factoring $9x^2-1$
$endgroup$
– J. W. Tanner
Mar 6 at 23:19
$begingroup$
Welcome to Math Stack Exchange. Please use MathJax. Consider factoring $9x^2-1$
$endgroup$
– J. W. Tanner
Mar 6 at 23:19
2
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
$begingroup$
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 Answer
1
active
oldest
votes
$begingroup$
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.
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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$begingroup$
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.
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.
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.
edited Mar 6 at 23:39
J. W. Tanner
3,8871320
answered Mar 6 at 23:20
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
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.
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
edited Mar 6 at 23:39
J. W. Tanner
3,8871320
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
answered Mar 6 at 23:20
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
add a comment |
add a comment |
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$begingroup$
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