How to find polynomial functions 3rd degree with no, one, two, three zeros(roots)?












0












$begingroup$


I must find 3rd degree Polynomial functions in R[x] with:



1) no roots



2) only one root



3) only two roots



4) only 3 roots



If the function has a root, then prove it. If not, then explain why.



My attempt:



We know, that the cubic function can have one, two or three roots.
But I really don't know, how I can find the polynomial functions.



1) Explanation



A 3rd polynomial function can not have no root because a polynomial function have at least one root. (Continous function)






linear-algebra polynomials roots







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    By root, do you mean only real roots, such as $-1, pi$ etc.?All polynomials of degree $n$ in one variable have exactly $n$ complex roots, counting multiplicity.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 15:43












  • $begingroup$
    What about $(x - a)^3$ and $(x-a)^2(x-b)$ and $(x-a)(x-b)(x-c)$ where $a, , b,,c$ are distinct numbers, where "numbers" is whatever type of numbers you need this for (rational numbers, real numbers, complex numbers, positive integers with exactly five prime divisors, etc.)?
    $endgroup$
    – Dave L. Renfro
    Dec 6 '18 at 15:57








  • 1




    $begingroup$
    For part 1, the continuity is not enough. Are you saying that there is no continuous function that has no roots? You also need to say that the function changes sign, since the limits at $+infty$ and $-infty$ have opposite sign. Since it's continuous, it means that it reaches all values between the limits, including $0$.
    $endgroup$
    – Andrei
    Dec 6 '18 at 16:05










  • $begingroup$
    Its important to clarify if you are counting multiplicity or not. No degree three polynomial has two real roots if you count multiplicity, but there are degree three polynomials with only two distinct real roots.
    $endgroup$
    – Eric
    Dec 6 '18 at 16:47
















0












$begingroup$


I must find 3rd degree Polynomial functions in R[x] with:



1) no roots



2) only one root



3) only two roots



4) only 3 roots



If the function has a root, then prove it. If not, then explain why.



My attempt:



We know, that the cubic function can have one, two or three roots.
But I really don't know, how I can find the polynomial functions.



1) Explanation



A 3rd polynomial function can not have no root because a polynomial function have at least one root. (Continous function)






linear-algebra polynomials roots







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    By root, do you mean only real roots, such as $-1, pi$ etc.?All polynomials of degree $n$ in one variable have exactly $n$ complex roots, counting multiplicity.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 15:43












  • $begingroup$
    What about $(x - a)^3$ and $(x-a)^2(x-b)$ and $(x-a)(x-b)(x-c)$ where $a, , b,,c$ are distinct numbers, where "numbers" is whatever type of numbers you need this for (rational numbers, real numbers, complex numbers, positive integers with exactly five prime divisors, etc.)?
    $endgroup$
    – Dave L. Renfro
    Dec 6 '18 at 15:57








  • 1




    $begingroup$
    For part 1, the continuity is not enough. Are you saying that there is no continuous function that has no roots? You also need to say that the function changes sign, since the limits at $+infty$ and $-infty$ have opposite sign. Since it's continuous, it means that it reaches all values between the limits, including $0$.
    $endgroup$
    – Andrei
    Dec 6 '18 at 16:05










  • $begingroup$
    Its important to clarify if you are counting multiplicity or not. No degree three polynomial has two real roots if you count multiplicity, but there are degree three polynomials with only two distinct real roots.
    $endgroup$
    – Eric
    Dec 6 '18 at 16:47














0












0








0





$begingroup$


I must find 3rd degree Polynomial functions in R[x] with:



1) no roots



2) only one root



3) only two roots



4) only 3 roots



If the function has a root, then prove it. If not, then explain why.



My attempt:



We know, that the cubic function can have one, two or three roots.
But I really don't know, how I can find the polynomial functions.



1) Explanation



A 3rd polynomial function can not have no root because a polynomial function have at least one root. (Continous function)






linear-algebra polynomials roots







share|cite|improve this question









$endgroup$




I must find 3rd degree Polynomial functions in R[x] with:



1) no roots



2) only one root



3) only two roots



4) only 3 roots



If the function has a root, then prove it. If not, then explain why.



My attempt:



We know, that the cubic function can have one, two or three roots.
But I really don't know, how I can find the polynomial functions.



1) Explanation



A 3rd polynomial function can not have no root because a polynomial function have at least one root. (Continous function)






linear-algebra polynomials roots




linear-algebra polynomials roots






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 15:41









MaLMaL

132




132








  • 1




    $begingroup$
    By root, do you mean only real roots, such as $-1, pi$ etc.?All polynomials of degree $n$ in one variable have exactly $n$ complex roots, counting multiplicity.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 15:43












  • $begingroup$
    What about $(x - a)^3$ and $(x-a)^2(x-b)$ and $(x-a)(x-b)(x-c)$ where $a, , b,,c$ are distinct numbers, where "numbers" is whatever type of numbers you need this for (rational numbers, real numbers, complex numbers, positive integers with exactly five prime divisors, etc.)?
    $endgroup$
    – Dave L. Renfro
    Dec 6 '18 at 15:57








  • 1




    $begingroup$
    For part 1, the continuity is not enough. Are you saying that there is no continuous function that has no roots? You also need to say that the function changes sign, since the limits at $+infty$ and $-infty$ have opposite sign. Since it's continuous, it means that it reaches all values between the limits, including $0$.
    $endgroup$
    – Andrei
    Dec 6 '18 at 16:05










  • $begingroup$
    Its important to clarify if you are counting multiplicity or not. No degree three polynomial has two real roots if you count multiplicity, but there are degree three polynomials with only two distinct real roots.
    $endgroup$
    – Eric
    Dec 6 '18 at 16:47














  • 1




    $begingroup$
    By root, do you mean only real roots, such as $-1, pi$ etc.?All polynomials of degree $n$ in one variable have exactly $n$ complex roots, counting multiplicity.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 15:43












  • $begingroup$
    What about $(x - a)^3$ and $(x-a)^2(x-b)$ and $(x-a)(x-b)(x-c)$ where $a, , b,,c$ are distinct numbers, where "numbers" is whatever type of numbers you need this for (rational numbers, real numbers, complex numbers, positive integers with exactly five prime divisors, etc.)?
    $endgroup$
    – Dave L. Renfro
    Dec 6 '18 at 15:57








  • 1




    $begingroup$
    For part 1, the continuity is not enough. Are you saying that there is no continuous function that has no roots? You also need to say that the function changes sign, since the limits at $+infty$ and $-infty$ have opposite sign. Since it's continuous, it means that it reaches all values between the limits, including $0$.
    $endgroup$
    – Andrei
    Dec 6 '18 at 16:05










  • $begingroup$
    Its important to clarify if you are counting multiplicity or not. No degree three polynomial has two real roots if you count multiplicity, but there are degree three polynomials with only two distinct real roots.
    $endgroup$
    – Eric
    Dec 6 '18 at 16:47








1




1




$begingroup$
By root, do you mean only real roots, such as $-1, pi$ etc.?All polynomials of degree $n$ in one variable have exactly $n$ complex roots, counting multiplicity.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 6 '18 at 15:43






$begingroup$
By root, do you mean only real roots, such as $-1, pi$ etc.?All polynomials of degree $n$ in one variable have exactly $n$ complex roots, counting multiplicity.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 6 '18 at 15:43














$begingroup$
What about $(x - a)^3$ and $(x-a)^2(x-b)$ and $(x-a)(x-b)(x-c)$ where $a, , b,,c$ are distinct numbers, where "numbers" is whatever type of numbers you need this for (rational numbers, real numbers, complex numbers, positive integers with exactly five prime divisors, etc.)?
$endgroup$
– Dave L. Renfro
Dec 6 '18 at 15:57






$begingroup$
What about $(x - a)^3$ and $(x-a)^2(x-b)$ and $(x-a)(x-b)(x-c)$ where $a, , b,,c$ are distinct numbers, where "numbers" is whatever type of numbers you need this for (rational numbers, real numbers, complex numbers, positive integers with exactly five prime divisors, etc.)?
$endgroup$
– Dave L. Renfro
Dec 6 '18 at 15:57






1




1




$begingroup$
For part 1, the continuity is not enough. Are you saying that there is no continuous function that has no roots? You also need to say that the function changes sign, since the limits at $+infty$ and $-infty$ have opposite sign. Since it's continuous, it means that it reaches all values between the limits, including $0$.
$endgroup$
– Andrei
Dec 6 '18 at 16:05




$begingroup$
For part 1, the continuity is not enough. Are you saying that there is no continuous function that has no roots? You also need to say that the function changes sign, since the limits at $+infty$ and $-infty$ have opposite sign. Since it's continuous, it means that it reaches all values between the limits, including $0$.
$endgroup$
– Andrei
Dec 6 '18 at 16:05












$begingroup$
Its important to clarify if you are counting multiplicity or not. No degree three polynomial has two real roots if you count multiplicity, but there are degree three polynomials with only two distinct real roots.
$endgroup$
– Eric
Dec 6 '18 at 16:47




$begingroup$
Its important to clarify if you are counting multiplicity or not. No degree three polynomial has two real roots if you count multiplicity, but there are degree three polynomials with only two distinct real roots.
$endgroup$
– Eric
Dec 6 '18 at 16:47










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Hint: make them in a standard form as $(x-x_1)(x-x_2)(x-x_3)$ where $x_1,x_2,x_3$ are the roots. For example for the case where we have $2$ zero roots we have $$x_1=x_2=0$$and choosing $x_3=1$ arbitrarily we obtain$$f(x)=x^2(x-1)$$






share|cite|improve this answer









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    $begingroup$

    Hint: make them in a standard form as $(x-x_1)(x-x_2)(x-x_3)$ where $x_1,x_2,x_3$ are the roots. For example for the case where we have $2$ zero roots we have $$x_1=x_2=0$$and choosing $x_3=1$ arbitrarily we obtain$$f(x)=x^2(x-1)$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint: make them in a standard form as $(x-x_1)(x-x_2)(x-x_3)$ where $x_1,x_2,x_3$ are the roots. For example for the case where we have $2$ zero roots we have $$x_1=x_2=0$$and choosing $x_3=1$ arbitrarily we obtain$$f(x)=x^2(x-1)$$






      share|cite|improve this answer









      $endgroup$
















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        0








        0





        $begingroup$

        Hint: make them in a standard form as $(x-x_1)(x-x_2)(x-x_3)$ where $x_1,x_2,x_3$ are the roots. For example for the case where we have $2$ zero roots we have $$x_1=x_2=0$$and choosing $x_3=1$ arbitrarily we obtain$$f(x)=x^2(x-1)$$






        share|cite|improve this answer









        $endgroup$



        Hint: make them in a standard form as $(x-x_1)(x-x_2)(x-x_3)$ where $x_1,x_2,x_3$ are the roots. For example for the case where we have $2$ zero roots we have $$x_1=x_2=0$$and choosing $x_3=1$ arbitrarily we obtain$$f(x)=x^2(x-1)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 20:19









        Mostafa AyazMostafa Ayaz

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