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Do Cubics always have one real root?

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Do Cubics always have one real root?

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1

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

Servaes

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asked 3 hours ago

user7971589user7971589

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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 | 

1

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

Servaes

27.8k34098

asked 3 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

share|cite|improve this question

edited 3 hours ago

Servaes

27.8k34098

asked 3 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.

$endgroup$

share|cite|improve this question

edited 3 hours ago

Servaes

27.8k34098

asked 3 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.

share|cite|improve this question

edited 3 hours ago

Servaes

27.8k34098

asked 3 hours ago

user7971589user7971589

82

New contributor

user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 3 hours ago

Servaes

27.8k34098

edited 3 hours ago

Servaes

27.8k34098

asked 3 hours ago

user7971589user7971589

82

New contributor

user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 3 hours ago

user7971589user7971589

82

2 Answers
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4

$begingroup$

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 $f(z) = (z-i)(z+i)(z-2i)$.

share|cite|improve this answer

answered 3 hours ago

RandallRandall

10.3k11230

$endgroup$

add a comment | 

3

$begingroup$

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 $x^3+i$.

share|cite|improve this answer

answered 3 hours ago

ServaesServaes

27.8k34098

$endgroup$

add a comment | 

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4

$begingroup$

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 $f(z) = (z-i)(z+i)(z-2i)$.

share|cite|improve this answer

answered 3 hours ago

RandallRandall

10.3k11230

$endgroup$

add a comment | 

4

$begingroup$

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 $f(z) = (z-i)(z+i)(z-2i)$.

share|cite|improve this answer

answered 3 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 $f(z) = (z-i)(z+i)(z-2i)$.

share|cite|improve this answer

answered 3 hours ago

RandallRandall

10.3k11230

$endgroup$

share|cite|improve this answer

answered 3 hours ago

RandallRandall

10.3k11230

answered 3 hours ago

RandallRandall

10.3k11230

answered 3 hours ago

RandallRandall

10.3k11230

3

$begingroup$

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 $x^3+i$.

share|cite|improve this answer

answered 3 hours ago

ServaesServaes

27.8k34098

$endgroup$

add a comment | 

3

$begingroup$

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 $x^3+i$.

share|cite|improve this answer

answered 3 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 $x^3+i$.

share|cite|improve this answer

answered 3 hours ago

ServaesServaes

27.8k34098

$endgroup$

share|cite|improve this answer

answered 3 hours ago

ServaesServaes

27.8k34098

answered 3 hours ago

ServaesServaes

27.8k34098

answered 3 hours ago

ServaesServaes

27.8k34098