Simple algebra question (find the LCM of the polynomials by factoring)
up vote
5
down vote
favorite
First of all, I am so embarrassed by this. I am tutoring this kid in math and this question came up: Find the LCM of $y^2 - 81$ and $9 - y$, which factor into $(y + 9)(y - 9)$ and $-(y - 9)$
My answer: $-(y+9)(y-9)$
Her book's answer: $(y+9)(y-9)$
My question: Why is the negative left off? Isn't it ($-1$) a factor of the second binomial?
algebra-precalculus
add a comment |
up vote
5
down vote
favorite
First of all, I am so embarrassed by this. I am tutoring this kid in math and this question came up: Find the LCM of $y^2 - 81$ and $9 - y$, which factor into $(y + 9)(y - 9)$ and $-(y - 9)$
My answer: $-(y+9)(y-9)$
Her book's answer: $(y+9)(y-9)$
My question: Why is the negative left off? Isn't it ($-1$) a factor of the second binomial?
algebra-precalculus
gcds & lcms in polynomial rings are defined only up to unit (invertible) factors. For polynomials (over fields) they usually normalized to be monic, i.e lead coef $= 1.,$ See here for more on such unit normalization.
– Bill Dubuque
Nov 21 at 2:02
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
First of all, I am so embarrassed by this. I am tutoring this kid in math and this question came up: Find the LCM of $y^2 - 81$ and $9 - y$, which factor into $(y + 9)(y - 9)$ and $-(y - 9)$
My answer: $-(y+9)(y-9)$
Her book's answer: $(y+9)(y-9)$
My question: Why is the negative left off? Isn't it ($-1$) a factor of the second binomial?
algebra-precalculus
First of all, I am so embarrassed by this. I am tutoring this kid in math and this question came up: Find the LCM of $y^2 - 81$ and $9 - y$, which factor into $(y + 9)(y - 9)$ and $-(y - 9)$
My answer: $-(y+9)(y-9)$
Her book's answer: $(y+9)(y-9)$
My question: Why is the negative left off? Isn't it ($-1$) a factor of the second binomial?
algebra-precalculus
algebra-precalculus
asked Nov 21 at 1:52
Ninosław Ciszewski
516411
516411
gcds & lcms in polynomial rings are defined only up to unit (invertible) factors. For polynomials (over fields) they usually normalized to be monic, i.e lead coef $= 1.,$ See here for more on such unit normalization.
– Bill Dubuque
Nov 21 at 2:02
add a comment |
gcds & lcms in polynomial rings are defined only up to unit (invertible) factors. For polynomials (over fields) they usually normalized to be monic, i.e lead coef $= 1.,$ See here for more on such unit normalization.
– Bill Dubuque
Nov 21 at 2:02
gcds & lcms in polynomial rings are defined only up to unit (invertible) factors. For polynomials (over fields) they usually normalized to be monic, i.e lead coef $= 1.,$ See here for more on such unit normalization.
– Bill Dubuque
Nov 21 at 2:02
gcds & lcms in polynomial rings are defined only up to unit (invertible) factors. For polynomials (over fields) they usually normalized to be monic, i.e lead coef $= 1.,$ See here for more on such unit normalization.
– Bill Dubuque
Nov 21 at 2:02
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
Both answers are valid. Any nonzero constant can be factored out of any polynomial.
Technically speaking, finding the $operatorname{lcm}$ of two elements is an action done in a ring. If it exists, the $operatorname{lcm}$ is unique up to multiplication of a unit of the ring, i.e. an invertible element.
In this case, all nonzero constants are units in a polynomial ring over a field (here $mathbb{R}$), so multiplying by a constant still gives an $operatorname{lcm}$. If we instead are considering these polynomials over $mathbb{Z}$, there's still no problem since $-1$ is a unit.
"all nonzero constants are units". Also better to restrict to domains vs. rings since divisibility theory in general rings is more complex, e.g. associates are no longer the same as unit multiples; furthermore the notions of associate and irreducible bifurcate into a few inequivalent notions in use.
– Bill Dubuque
Nov 21 at 2:32
Good point. Could you edit to add the nuance for rings vs domains? My ring theory's a bit rusty.
– Sambo
Nov 21 at 3:13
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Both answers are valid. Any nonzero constant can be factored out of any polynomial.
Technically speaking, finding the $operatorname{lcm}$ of two elements is an action done in a ring. If it exists, the $operatorname{lcm}$ is unique up to multiplication of a unit of the ring, i.e. an invertible element.
In this case, all nonzero constants are units in a polynomial ring over a field (here $mathbb{R}$), so multiplying by a constant still gives an $operatorname{lcm}$. If we instead are considering these polynomials over $mathbb{Z}$, there's still no problem since $-1$ is a unit.
"all nonzero constants are units". Also better to restrict to domains vs. rings since divisibility theory in general rings is more complex, e.g. associates are no longer the same as unit multiples; furthermore the notions of associate and irreducible bifurcate into a few inequivalent notions in use.
– Bill Dubuque
Nov 21 at 2:32
Good point. Could you edit to add the nuance for rings vs domains? My ring theory's a bit rusty.
– Sambo
Nov 21 at 3:13
add a comment |
up vote
4
down vote
Both answers are valid. Any nonzero constant can be factored out of any polynomial.
Technically speaking, finding the $operatorname{lcm}$ of two elements is an action done in a ring. If it exists, the $operatorname{lcm}$ is unique up to multiplication of a unit of the ring, i.e. an invertible element.
In this case, all nonzero constants are units in a polynomial ring over a field (here $mathbb{R}$), so multiplying by a constant still gives an $operatorname{lcm}$. If we instead are considering these polynomials over $mathbb{Z}$, there's still no problem since $-1$ is a unit.
"all nonzero constants are units". Also better to restrict to domains vs. rings since divisibility theory in general rings is more complex, e.g. associates are no longer the same as unit multiples; furthermore the notions of associate and irreducible bifurcate into a few inequivalent notions in use.
– Bill Dubuque
Nov 21 at 2:32
Good point. Could you edit to add the nuance for rings vs domains? My ring theory's a bit rusty.
– Sambo
Nov 21 at 3:13
add a comment |
up vote
4
down vote
up vote
4
down vote
Both answers are valid. Any nonzero constant can be factored out of any polynomial.
Technically speaking, finding the $operatorname{lcm}$ of two elements is an action done in a ring. If it exists, the $operatorname{lcm}$ is unique up to multiplication of a unit of the ring, i.e. an invertible element.
In this case, all nonzero constants are units in a polynomial ring over a field (here $mathbb{R}$), so multiplying by a constant still gives an $operatorname{lcm}$. If we instead are considering these polynomials over $mathbb{Z}$, there's still no problem since $-1$ is a unit.
Both answers are valid. Any nonzero constant can be factored out of any polynomial.
Technically speaking, finding the $operatorname{lcm}$ of two elements is an action done in a ring. If it exists, the $operatorname{lcm}$ is unique up to multiplication of a unit of the ring, i.e. an invertible element.
In this case, all nonzero constants are units in a polynomial ring over a field (here $mathbb{R}$), so multiplying by a constant still gives an $operatorname{lcm}$. If we instead are considering these polynomials over $mathbb{Z}$, there's still no problem since $-1$ is a unit.
edited Nov 21 at 3:13
answered Nov 21 at 2:05
Sambo
2,1112532
2,1112532
"all nonzero constants are units". Also better to restrict to domains vs. rings since divisibility theory in general rings is more complex, e.g. associates are no longer the same as unit multiples; furthermore the notions of associate and irreducible bifurcate into a few inequivalent notions in use.
– Bill Dubuque
Nov 21 at 2:32
Good point. Could you edit to add the nuance for rings vs domains? My ring theory's a bit rusty.
– Sambo
Nov 21 at 3:13
add a comment |
"all nonzero constants are units". Also better to restrict to domains vs. rings since divisibility theory in general rings is more complex, e.g. associates are no longer the same as unit multiples; furthermore the notions of associate and irreducible bifurcate into a few inequivalent notions in use.
– Bill Dubuque
Nov 21 at 2:32
Good point. Could you edit to add the nuance for rings vs domains? My ring theory's a bit rusty.
– Sambo
Nov 21 at 3:13
"all nonzero constants are units". Also better to restrict to domains vs. rings since divisibility theory in general rings is more complex, e.g. associates are no longer the same as unit multiples; furthermore the notions of associate and irreducible bifurcate into a few inequivalent notions in use.
– Bill Dubuque
Nov 21 at 2:32
"all nonzero constants are units". Also better to restrict to domains vs. rings since divisibility theory in general rings is more complex, e.g. associates are no longer the same as unit multiples; furthermore the notions of associate and irreducible bifurcate into a few inequivalent notions in use.
– Bill Dubuque
Nov 21 at 2:32
Good point. Could you edit to add the nuance for rings vs domains? My ring theory's a bit rusty.
– Sambo
Nov 21 at 3:13
Good point. Could you edit to add the nuance for rings vs domains? My ring theory's a bit rusty.
– Sambo
Nov 21 at 3:13
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007151%2fsimple-algebra-question-find-the-lcm-of-the-polynomials-by-factoring%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
gcds & lcms in polynomial rings are defined only up to unit (invertible) factors. For polynomials (over fields) they usually normalized to be monic, i.e lead coef $= 1.,$ See here for more on such unit normalization.
– Bill Dubuque
Nov 21 at 2:02