Mixing
Simple algebra question (find the LCM of the polynomials by factoring)
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
asked Nov 21 '18 at 1:52
Ninosław Ciszewski
521412
add a comment |
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
asked Nov 21 '18 at 1:52
Ninosław Ciszewski
521412
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 '18 at 2:02
add a comment |
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
asked Nov 21 '18 at 1:52
Ninosław Ciszewski
521412
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 '18 at 1:52
Ninosław Ciszewski
521412
asked Nov 21 '18 at 1:52
Ninosław Ciszewski
521412
asked Nov 21 '18 at 1:52
Ninosław Ciszewski
521412
asked Nov 21 '18 at 1:52
Ninosław Ciszewski
521412
asked Nov 21 '18 at 1:52
Ninosław Ciszewski
521412
521412
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 '18 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 '18 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 '18 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 '18 at 2:02
add a comment |
1 Answer
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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.
answered Nov 21 '18 at 2:05
Sambo
2,1382532
"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 '18 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 '18 at 3:13
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 21 '18 at 2:05
Sambo
2,1382532
"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 '18 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 '18 at 3:13
add a comment |
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.
answered Nov 21 '18 at 2:05
Sambo
2,1382532
"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 '18 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 '18 at 3:13
add a comment |
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.
answered Nov 21 '18 at 2:05
Sambo
2,1382532
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.
answered Nov 21 '18 at 2:05
Sambo
2,1382532
edited Nov 21 '18 at 3:13
answered Nov 21 '18 at 2:05
Sambo
2,1382532
answered Nov 21 '18 at 2:05
Sambo
2,1382532
answered Nov 21 '18 at 2:05
Sambo
2,1382532
2,1382532
"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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 3:13
add a comment |
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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 '18 at 2:02