Do Cubics always have one real root?Does the resolvent cubic of the quartic equation always have at least 1 positive real rootCubic polynomial with 1 real root and 2 complex conjugated roots (real coefficients)Complex Conjugate roots with non real coefficientsFind coefficients of a cubic function with imaginary rootCubic Function with two roots and its Derivative function with one rootWhat's the easiest way to solve the cubic $3x^3-13x^2+3x-13$Apart from the Fundamental Theorem of Algebra and Descartes Rule of Signs, are there any other ways to determine the nature of roots of a polynomial?How many real roots can a cubic equation $x^3 + bx^2 + cx + d = 0$ have?How to tell root multiplicity from complex rootsProved that cubic equation w/ real coefficients always has 2 complex conjugate roots but that's clearly not the case.
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Do Cubics always have one real root?
Does the resolvent cubic of the quartic equation always have at least 1 positive real rootCubic polynomial with 1 real root and 2 complex conjugated roots (real coefficients)Complex Conjugate roots with non real coefficientsFind coefficients of a cubic function with imaginary rootCubic Function with two roots and its Derivative function with one rootWhat's the easiest way to solve the cubic $3x^3-13x^2+3x-13$Apart from the Fundamental Theorem of Algebra and Descartes Rule of Signs, are there any other ways to determine the nature of roots of a polynomial?How many real roots can a cubic equation $x^3 + bx^2 + cx + d = 0$ have?How to tell root multiplicity from complex rootsProved that cubic equation w/ real coefficients always has 2 complex conjugate roots but that's clearly not the case.
$begingroup$
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 5 hours ago
Servaes
27.8k34098
asked 5 hours ago
user7971589user7971589
112
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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 5 hours ago
Servaes
27.8k34098
asked 5 hours ago
user7971589user7971589
112
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 5 hours ago
Servaes
27.8k34098
asked 5 hours ago
user7971589user7971589
112
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 5 hours ago
Servaes
27.8k34098
asked 5 hours ago
user7971589user7971589
112
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 5 hours ago
Servaes
27.8k34098
asked 5 hours ago
user7971589user7971589
112
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 5 hours ago
Servaes
27.8k34098
edited 5 hours ago
Servaes
27.8k34098
edited 5 hours ago
Servaes
27.8k34098
27.8k34098
asked 5 hours ago
user7971589user7971589
112
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 5 hours ago
user7971589user7971589
112
asked 5 hours ago
user7971589user7971589
112
112
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 |
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 5 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 5 hours ago
ServaesServaes
27.8k34098
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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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 5 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 5 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 5 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 5 hours ago
RandallRandall
10.3k11230
answered 5 hours ago
RandallRandall
10.3k11230
answered 5 hours ago
RandallRandall
10.3k11230
answered 5 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 5 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 5 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 5 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 5 hours ago
ServaesServaes
27.8k34098
answered 5 hours ago
ServaesServaes
27.8k34098
answered 5 hours ago
ServaesServaes
27.8k34098
answered 5 hours ago
ServaesServaes
27.8k34098
27.8k34098
add a comment |
add a comment |
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