Root(zero) of a polynomial
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How is the Root of a polynomial defined? I found on Wikipedia's page written: "A root of a polynomial is a zero of the corresponding polynomial function." Or the URL:https://en.m.wikipedia.org/wiki/Zero_of_a_function
Now if I want to find the root of polynomial 9a-2b then that would be it's corresponding function?
polynomials
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add a comment |
$begingroup$
How is the Root of a polynomial defined? I found on Wikipedia's page written: "A root of a polynomial is a zero of the corresponding polynomial function." Or the URL:https://en.m.wikipedia.org/wiki/Zero_of_a_function
Now if I want to find the root of polynomial 9a-2b then that would be it's corresponding function?
polynomials
$endgroup$
add a comment |
$begingroup$
How is the Root of a polynomial defined? I found on Wikipedia's page written: "A root of a polynomial is a zero of the corresponding polynomial function." Or the URL:https://en.m.wikipedia.org/wiki/Zero_of_a_function
Now if I want to find the root of polynomial 9a-2b then that would be it's corresponding function?
polynomials
$endgroup$
How is the Root of a polynomial defined? I found on Wikipedia's page written: "A root of a polynomial is a zero of the corresponding polynomial function." Or the URL:https://en.m.wikipedia.org/wiki/Zero_of_a_function
Now if I want to find the root of polynomial 9a-2b then that would be it's corresponding function?
polynomials
polynomials
asked Dec 30 '18 at 8:57
user629353
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$
answered Dec 30 '18 at 9:00
José Carlos SantosJosé Carlos Santos
164k22131234
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$begingroup$
Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
$endgroup$
– user629353
Dec 30 '18 at 9:02
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It's the functions that maps the pair $(a,b)$ into $9a-2b$.
$endgroup$
– José Carlos Santos
Dec 30 '18 at 9:05
add a comment |
$begingroup$
A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.
answered Dec 30 '18 at 9:05
Amey JoshiAmey Joshi
604310
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add a comment |
$begingroup$
$9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.
If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.
answered Dec 30 '18 at 9:29
Henno BrandsmaHenno Brandsma
111k348118
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$
answered Dec 30 '18 at 9:00
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
$endgroup$
– user629353
Dec 30 '18 at 9:02
$begingroup$
It's the functions that maps the pair $(a,b)$ into $9a-2b$.
$endgroup$
– José Carlos Santos
Dec 30 '18 at 9:05
add a comment |
$begingroup$
If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$
answered Dec 30 '18 at 9:00
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
$endgroup$
– user629353
Dec 30 '18 at 9:02
$begingroup$
It's the functions that maps the pair $(a,b)$ into $9a-2b$.
$endgroup$
– José Carlos Santos
Dec 30 '18 at 9:05
add a comment |
$begingroup$
If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$
answered Dec 30 '18 at 9:00
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$
answered Dec 30 '18 at 9:00
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 30 '18 at 9:00
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 30 '18 at 9:00
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 30 '18 at 9:00
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
$endgroup$
– user629353
Dec 30 '18 at 9:02
$begingroup$
It's the functions that maps the pair $(a,b)$ into $9a-2b$.
$endgroup$
– José Carlos Santos
Dec 30 '18 at 9:05
add a comment |
$begingroup$
Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
$endgroup$
– user629353
Dec 30 '18 at 9:02
$begingroup$
It's the functions that maps the pair $(a,b)$ into $9a-2b$.
$endgroup$
– José Carlos Santos
Dec 30 '18 at 9:05
$begingroup$
Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
$endgroup$
– user629353
Dec 30 '18 at 9:02
$begingroup$
Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
$endgroup$
– user629353
Dec 30 '18 at 9:02
$begingroup$
It's the functions that maps the pair $(a,b)$ into $9a-2b$.
$endgroup$
– José Carlos Santos
Dec 30 '18 at 9:05
$begingroup$
It's the functions that maps the pair $(a,b)$ into $9a-2b$.
$endgroup$
– José Carlos Santos
Dec 30 '18 at 9:05
add a comment |
$begingroup$
A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.
answered Dec 30 '18 at 9:05
Amey JoshiAmey Joshi
604310
$endgroup$
add a comment |
$begingroup$
A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.
answered Dec 30 '18 at 9:05
Amey JoshiAmey Joshi
604310
$endgroup$
add a comment |
$begingroup$
A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.
answered Dec 30 '18 at 9:05
Amey JoshiAmey Joshi
604310
$endgroup$
A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.
answered Dec 30 '18 at 9:05
Amey JoshiAmey Joshi
604310
answered Dec 30 '18 at 9:05
Amey JoshiAmey Joshi
604310
answered Dec 30 '18 at 9:05
Amey JoshiAmey Joshi
604310
answered Dec 30 '18 at 9:05
Amey JoshiAmey Joshi
604310
604310
add a comment |
add a comment |
$begingroup$
$9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.
If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.
answered Dec 30 '18 at 9:29
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
add a comment |
$begingroup$
$9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.
If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.
answered Dec 30 '18 at 9:29
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
add a comment |
$begingroup$
$9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.
If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.
answered Dec 30 '18 at 9:29
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.
If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.
answered Dec 30 '18 at 9:29
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 30 '18 at 9:29
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 30 '18 at 9:29
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 30 '18 at 9:29
Henno BrandsmaHenno Brandsma
111k348118
111k348118
add a comment |
add a comment |
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