Root of polynomial in $mathbb{Z}_{13}$
$begingroup$
I want to find the roots of the polynomial $2T^7 + 3T^6 + 6T^5 + 11T^3 + 9T + 8$ in the field $mathbb{Z}_{13}$.
I did a polynomial division by $3T^4 + 11T^2 + 2T + 9$ which resulted in:
$(3T^4 + 11T^2 + 2T + 9)(5T^3 + T^2 + T + 6) + (5T^3 + T^2+ T + 6)$
This can be rewritten to
$(5T^3 + 3T^2 + T + 6)(3T^4 + 11T^2 + 2T +10)$
I tried to find a root for the term $(5T^3 + 3T^2 + T + 6)$ by just inserting 1 to 12 for $T$, but it was not valid in any case.
For the term $(3T^4 + 11T^2 + 2T +10)$ I do not have any clue at all.
How can I find a root here?
polynomials finite-groups
asked Jan 4 at 7:48
HellstormHellstorm
182
$endgroup$
|
show 1 more comment
$begingroup$
I want to find the roots of the polynomial $2T^7 + 3T^6 + 6T^5 + 11T^3 + 9T + 8$ in the field $mathbb{Z}_{13}$.
I did a polynomial division by $3T^4 + 11T^2 + 2T + 9$ which resulted in:
$(3T^4 + 11T^2 + 2T + 9)(5T^3 + T^2 + T + 6) + (5T^3 + T^2+ T + 6)$
This can be rewritten to
$(5T^3 + 3T^2 + T + 6)(3T^4 + 11T^2 + 2T +10)$
I tried to find a root for the term $(5T^3 + 3T^2 + T + 6)$ by just inserting 1 to 12 for $T$, but it was not valid in any case.
For the term $(3T^4 + 11T^2 + 2T +10)$ I do not have any clue at all.
How can I find a root here?
polynomials finite-groups
asked Jan 4 at 7:48
HellstormHellstorm
182
$endgroup$
1
$begingroup$
Surely $T=1$ is a root of the quartic?
$endgroup$
– ancientmathematician
Jan 4 at 7:54
1
$begingroup$
$T=1$ is a root of the original polynomial. Maybe long divide the original polynomial by $(T-1)$ and see how it goes from there.
$endgroup$
– MAX
Jan 4 at 8:01
1
$begingroup$
How did you come ups with this factorization? Why have you divided by $3T^4+11T^2+2T+9$ and not by something else? Anyway, once you have this factorization, the roots of the original polynomial are precisely the roots of any of the two factors, you can find them just by inspection (plugging in all $13$ elements of $mathbb Z_{13}$.
$endgroup$
– W-t-P
Jan 4 at 8:04
1
$begingroup$
Just substitute $T=alpha$ for each element $alpha$ in ${Bbb Z}_{13}$.
$endgroup$
– Wuestenfux
Jan 4 at 8:53
$begingroup$
@W-t-P The task was to divide it by $3T^4 + 11T^2 + 2T + 9$. Is there a trick for checking if there are roots except for plugging everything into it? It's still manageable with only 13 elements, but still takes some time.
$endgroup$
– Hellstorm
Jan 4 at 12:22
|
show 1 more comment
$begingroup$
I want to find the roots of the polynomial $2T^7 + 3T^6 + 6T^5 + 11T^3 + 9T + 8$ in the field $mathbb{Z}_{13}$.
I did a polynomial division by $3T^4 + 11T^2 + 2T + 9$ which resulted in:
$(3T^4 + 11T^2 + 2T + 9)(5T^3 + T^2 + T + 6) + (5T^3 + T^2+ T + 6)$
This can be rewritten to
$(5T^3 + 3T^2 + T + 6)(3T^4 + 11T^2 + 2T +10)$
I tried to find a root for the term $(5T^3 + 3T^2 + T + 6)$ by just inserting 1 to 12 for $T$, but it was not valid in any case.
For the term $(3T^4 + 11T^2 + 2T +10)$ I do not have any clue at all.
How can I find a root here?
polynomials finite-groups
asked Jan 4 at 7:48
HellstormHellstorm
182
$endgroup$
I want to find the roots of the polynomial $2T^7 + 3T^6 + 6T^5 + 11T^3 + 9T + 8$ in the field $mathbb{Z}_{13}$.
I did a polynomial division by $3T^4 + 11T^2 + 2T + 9$ which resulted in:
$(3T^4 + 11T^2 + 2T + 9)(5T^3 + T^2 + T + 6) + (5T^3 + T^2+ T + 6)$
This can be rewritten to
$(5T^3 + 3T^2 + T + 6)(3T^4 + 11T^2 + 2T +10)$
I tried to find a root for the term $(5T^3 + 3T^2 + T + 6)$ by just inserting 1 to 12 for $T$, but it was not valid in any case.
For the term $(3T^4 + 11T^2 + 2T +10)$ I do not have any clue at all.
How can I find a root here?
polynomials finite-groups
polynomials finite-groups
asked Jan 4 at 7:48
HellstormHellstorm
182
asked Jan 4 at 7:48
HellstormHellstorm
182
asked Jan 4 at 7:48
HellstormHellstorm
182
asked Jan 4 at 7:48
HellstormHellstorm
182
asked Jan 4 at 7:48
HellstormHellstorm
182
182
1
$begingroup$
Surely $T=1$ is a root of the quartic?
$endgroup$
– ancientmathematician
Jan 4 at 7:54
1
$begingroup$
$T=1$ is a root of the original polynomial. Maybe long divide the original polynomial by $(T-1)$ and see how it goes from there.
$endgroup$
– MAX
Jan 4 at 8:01
1
$begingroup$
How did you come ups with this factorization? Why have you divided by $3T^4+11T^2+2T+9$ and not by something else? Anyway, once you have this factorization, the roots of the original polynomial are precisely the roots of any of the two factors, you can find them just by inspection (plugging in all $13$ elements of $mathbb Z_{13}$.
$endgroup$
– W-t-P
Jan 4 at 8:04
1
$begingroup$
Just substitute $T=alpha$ for each element $alpha$ in ${Bbb Z}_{13}$.
$endgroup$
– Wuestenfux
Jan 4 at 8:53
$begingroup$
@W-t-P The task was to divide it by $3T^4 + 11T^2 + 2T + 9$. Is there a trick for checking if there are roots except for plugging everything into it? It's still manageable with only 13 elements, but still takes some time.
$endgroup$
– Hellstorm
Jan 4 at 12:22
|
show 1 more comment
1
$begingroup$
Surely $T=1$ is a root of the quartic?
$endgroup$
– ancientmathematician
Jan 4 at 7:54
1
$begingroup$
$T=1$ is a root of the original polynomial. Maybe long divide the original polynomial by $(T-1)$ and see how it goes from there.
$endgroup$
– MAX
Jan 4 at 8:01
1
$begingroup$
How did you come ups with this factorization? Why have you divided by $3T^4+11T^2+2T+9$ and not by something else? Anyway, once you have this factorization, the roots of the original polynomial are precisely the roots of any of the two factors, you can find them just by inspection (plugging in all $13$ elements of $mathbb Z_{13}$.
$endgroup$
– W-t-P
Jan 4 at 8:04
1
$begingroup$
Just substitute $T=alpha$ for each element $alpha$ in ${Bbb Z}_{13}$.
$endgroup$
– Wuestenfux
Jan 4 at 8:53
$begingroup$
@W-t-P The task was to divide it by $3T^4 + 11T^2 + 2T + 9$. Is there a trick for checking if there are roots except for plugging everything into it? It's still manageable with only 13 elements, but still takes some time.
$endgroup$
– Hellstorm
Jan 4 at 12:22
1
1
$begingroup$
Surely $T=1$ is a root of the quartic?
$endgroup$
– ancientmathematician
Jan 4 at 7:54
$begingroup$
Surely $T=1$ is a root of the quartic?
$endgroup$
– ancientmathematician
Jan 4 at 7:54
1
1
$begingroup$
$T=1$ is a root of the original polynomial. Maybe long divide the original polynomial by $(T-1)$ and see how it goes from there.
$endgroup$
– MAX
Jan 4 at 8:01
$begingroup$
$T=1$ is a root of the original polynomial. Maybe long divide the original polynomial by $(T-1)$ and see how it goes from there.
$endgroup$
– MAX
Jan 4 at 8:01
1
1
$begingroup$
How did you come ups with this factorization? Why have you divided by $3T^4+11T^2+2T+9$ and not by something else? Anyway, once you have this factorization, the roots of the original polynomial are precisely the roots of any of the two factors, you can find them just by inspection (plugging in all $13$ elements of $mathbb Z_{13}$.
$endgroup$
– W-t-P
Jan 4 at 8:04
$begingroup$
How did you come ups with this factorization? Why have you divided by $3T^4+11T^2+2T+9$ and not by something else? Anyway, once you have this factorization, the roots of the original polynomial are precisely the roots of any of the two factors, you can find them just by inspection (plugging in all $13$ elements of $mathbb Z_{13}$.
$endgroup$
– W-t-P
Jan 4 at 8:04
1
1
$begingroup$
Just substitute $T=alpha$ for each element $alpha$ in ${Bbb Z}_{13}$.
$endgroup$
– Wuestenfux
Jan 4 at 8:53
$begingroup$
Just substitute $T=alpha$ for each element $alpha$ in ${Bbb Z}_{13}$.
$endgroup$
– Wuestenfux
Jan 4 at 8:53
$begingroup$
@W-t-P The task was to divide it by $3T^4 + 11T^2 + 2T + 9$. Is there a trick for checking if there are roots except for plugging everything into it? It's still manageable with only 13 elements, but still takes some time.
$endgroup$
– Hellstorm
Jan 4 at 12:22
$begingroup$
@W-t-P The task was to divide it by $3T^4 + 11T^2 + 2T + 9$. Is there a trick for checking if there are roots except for plugging everything into it? It's still manageable with only 13 elements, but still takes some time.
$endgroup$
– Hellstorm
Jan 4 at 12:22
|
show 1 more comment
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1
$begingroup$
Surely $T=1$ is a root of the quartic?
$endgroup$
– ancientmathematician
Jan 4 at 7:54
1
$begingroup$
$T=1$ is a root of the original polynomial. Maybe long divide the original polynomial by $(T-1)$ and see how it goes from there.
$endgroup$
– MAX
Jan 4 at 8:01
1
$begingroup$
How did you come ups with this factorization? Why have you divided by $3T^4+11T^2+2T+9$ and not by something else? Anyway, once you have this factorization, the roots of the original polynomial are precisely the roots of any of the two factors, you can find them just by inspection (plugging in all $13$ elements of $mathbb Z_{13}$.
$endgroup$
– W-t-P
Jan 4 at 8:04
1
$begingroup$
Just substitute $T=alpha$ for each element $alpha$ in ${Bbb Z}_{13}$.
$endgroup$
– Wuestenfux
Jan 4 at 8:53
$begingroup$
@W-t-P The task was to divide it by $3T^4 + 11T^2 + 2T + 9$. Is there a trick for checking if there are roots except for plugging everything into it? It's still manageable with only 13 elements, but still takes some time.
$endgroup$
– Hellstorm
Jan 4 at 12:22