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Do Cubics always have one real root?
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$begingroup$
I've seen a few conflicting pieces of information online.
So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?
At very least could you give me a counter example? a cubic with no real roots
polynomials complex-numbers roots real-numbers
edited 2 hours ago
Servaes
27.8k34098
asked 2 hours ago
user7971589user7971589
82
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user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I've seen a few conflicting pieces of information online.
So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?
At very least could you give me a counter example? a cubic with no real roots
polynomials complex-numbers roots real-numbers
edited 2 hours ago
Servaes
27.8k34098
asked 2 hours ago
user7971589user7971589
82
New contributor
user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I've seen a few conflicting pieces of information online.
So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?
At very least could you give me a counter example? a cubic with no real roots
polynomials complex-numbers roots real-numbers
edited 2 hours ago
Servaes
27.8k34098
asked 2 hours ago
user7971589user7971589
82
New contributor
user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I've seen a few conflicting pieces of information online.
So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?
At very least could you give me a counter example? a cubic with no real roots
polynomials complex-numbers roots real-numbers
polynomials complex-numbers roots real-numbers
edited 2 hours ago
Servaes
27.8k34098
asked 2 hours ago
user7971589user7971589
82
New contributor
user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Servaes
27.8k34098
asked 2 hours ago
user7971589user7971589
82
New contributor
user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Servaes
27.8k34098
edited 2 hours ago
Servaes
27.8k34098
edited 2 hours ago
Servaes
27.8k34098
27.8k34098
asked 2 hours ago
user7971589user7971589
82
New contributor
user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
user7971589user7971589
82
asked 2 hours ago
user7971589user7971589
82
82
New contributor
user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
2 Answers
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$begingroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered 2 hours ago
RandallRandall
10.3k11230
$endgroup$
add a comment |
$begingroup$
As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered 2 hours ago
ServaesServaes
27.8k34098
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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$begingroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered 2 hours ago
RandallRandall
10.3k11230
$endgroup$
add a comment |
$begingroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered 2 hours ago
RandallRandall
10.3k11230
$endgroup$
add a comment |
$begingroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered 2 hours ago
RandallRandall
10.3k11230
$endgroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered 2 hours ago
RandallRandall
10.3k11230
answered 2 hours ago
RandallRandall
10.3k11230
answered 2 hours ago
RandallRandall
10.3k11230
answered 2 hours ago
RandallRandall
10.3k11230
10.3k11230
add a comment |
add a comment |
$begingroup$
As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered 2 hours ago
ServaesServaes
27.8k34098
$endgroup$
add a comment |
$begingroup$
As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered 2 hours ago
ServaesServaes
27.8k34098
$endgroup$
add a comment |
$begingroup$
As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered 2 hours ago
ServaesServaes
27.8k34098
$endgroup$
As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered 2 hours ago
ServaesServaes
27.8k34098
answered 2 hours ago
ServaesServaes
27.8k34098
answered 2 hours ago
ServaesServaes
27.8k34098
answered 2 hours ago
ServaesServaes
27.8k34098
27.8k34098
add a comment |
add a comment |
user7971589 is a new contributor. Be nice, and check out our Code of Conduct.
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