Finding real coefficients of polynomial with complex roots

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I have been tasked with finding the just real coefficients of a polynomial with the roots:



$z_0=10$



$z_1=3-i$



$z_2=-8+2i$



Writing the polynomial as factors...



$(z-10)(z-3+i)(z+8-2i)$



..obviously does not work since this results in complex coefficients.



Can anyone help me out here?






algebra-precalculus polynomials complex-numbers







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  • “the real coefficients of a polynomial” or “a polynomial with only real coefficients”?
    – Martin R
    8 hours ago










  • Only real coefficients.
    – Boris Grunwald
    8 hours ago














up vote
3
down vote

favorite












I have been tasked with finding the just real coefficients of a polynomial with the roots:



$z_0=10$



$z_1=3-i$



$z_2=-8+2i$



Writing the polynomial as factors...



$(z-10)(z-3+i)(z+8-2i)$



..obviously does not work since this results in complex coefficients.



Can anyone help me out here?






algebra-precalculus polynomials complex-numbers







share|cite|improve this question























  • “the real coefficients of a polynomial” or “a polynomial with only real coefficients”?
    – Martin R
    8 hours ago










  • Only real coefficients.
    – Boris Grunwald
    8 hours ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I have been tasked with finding the just real coefficients of a polynomial with the roots:



$z_0=10$



$z_1=3-i$



$z_2=-8+2i$



Writing the polynomial as factors...



$(z-10)(z-3+i)(z+8-2i)$



..obviously does not work since this results in complex coefficients.



Can anyone help me out here?






algebra-precalculus polynomials complex-numbers







share|cite|improve this question















I have been tasked with finding the just real coefficients of a polynomial with the roots:



$z_0=10$



$z_1=3-i$



$z_2=-8+2i$



Writing the polynomial as factors...



$(z-10)(z-3+i)(z+8-2i)$



..obviously does not work since this results in complex coefficients.



Can anyone help me out here?






algebra-precalculus polynomials complex-numbers




algebra-precalculus polynomials complex-numbers






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









greedoid

31.1k94287




31.1k94287










asked 8 hours ago









Boris Grunwald

877




877











  • “the real coefficients of a polynomial” or “a polynomial with only real coefficients”?
    – Martin R
    8 hours ago










  • Only real coefficients.
    – Boris Grunwald
    8 hours ago
















  • “the real coefficients of a polynomial” or “a polynomial with only real coefficients”?
    – Martin R
    8 hours ago










  • Only real coefficients.
    – Boris Grunwald
    8 hours ago















“the real coefficients of a polynomial” or “a polynomial with only real coefficients”?
– Martin R
8 hours ago




“the real coefficients of a polynomial” or “a polynomial with only real coefficients”?
– Martin R
8 hours ago












Only real coefficients.
– Boris Grunwald
8 hours ago




Only real coefficients.
– Boris Grunwald
8 hours ago










2 Answers
2






active

oldest

votes

















up vote
5
down vote



accepted










There is not enough information, say the degree of a given polynomial.



Anyway, since it has (only) real coeficients if $a+bi$ is one root, then $a-bi$ is also a root of the polynomial.



So this polynomial is divisible with:



$$q(z)=(z-10)colorred(z-3+i)(z-3-i)colorblue(z+8-2i)(z+8+2i) $$
$$= (z-10)(z^2-6z+10)(z^2+16z+68)$$






share|cite|improve this answer






















  • Thanks, I think this makes sense.
    – Boris Grunwald
    8 hours ago










  • And sorry, it was the smallest possible degree
    – Boris Grunwald
    8 hours ago

















up vote
3
down vote













Polynomial with only real coefficients is
$$( z-10) , ( z-2i+8),(z+2 i+8) , ( z-i-3) , ( z+i-3)\=
z^5-118 z^3-68 z^2+3160 z-6800
$$






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    There is not enough information, say the degree of a given polynomial.



    Anyway, since it has (only) real coeficients if $a+bi$ is one root, then $a-bi$ is also a root of the polynomial.



    So this polynomial is divisible with:



    $$q(z)=(z-10)colorred(z-3+i)(z-3-i)colorblue(z+8-2i)(z+8+2i) $$
    $$= (z-10)(z^2-6z+10)(z^2+16z+68)$$






    share|cite|improve this answer






















    • Thanks, I think this makes sense.
      – Boris Grunwald
      8 hours ago










    • And sorry, it was the smallest possible degree
      – Boris Grunwald
      8 hours ago














    up vote
    5
    down vote



    accepted










    There is not enough information, say the degree of a given polynomial.



    Anyway, since it has (only) real coeficients if $a+bi$ is one root, then $a-bi$ is also a root of the polynomial.



    So this polynomial is divisible with:



    $$q(z)=(z-10)colorred(z-3+i)(z-3-i)colorblue(z+8-2i)(z+8+2i) $$
    $$= (z-10)(z^2-6z+10)(z^2+16z+68)$$






    share|cite|improve this answer






















    • Thanks, I think this makes sense.
      – Boris Grunwald
      8 hours ago










    • And sorry, it was the smallest possible degree
      – Boris Grunwald
      8 hours ago












    up vote
    5
    down vote



    accepted







    up vote
    5
    down vote



    accepted






    There is not enough information, say the degree of a given polynomial.



    Anyway, since it has (only) real coeficients if $a+bi$ is one root, then $a-bi$ is also a root of the polynomial.



    So this polynomial is divisible with:



    $$q(z)=(z-10)colorred(z-3+i)(z-3-i)colorblue(z+8-2i)(z+8+2i) $$
    $$= (z-10)(z^2-6z+10)(z^2+16z+68)$$






    share|cite|improve this answer














    There is not enough information, say the degree of a given polynomial.



    Anyway, since it has (only) real coeficients if $a+bi$ is one root, then $a-bi$ is also a root of the polynomial.



    So this polynomial is divisible with:



    $$q(z)=(z-10)colorred(z-3+i)(z-3-i)colorblue(z+8-2i)(z+8+2i) $$
    $$= (z-10)(z^2-6z+10)(z^2+16z+68)$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 8 hours ago

























    answered 8 hours ago









    greedoid

    31.1k94287




    31.1k94287











    • Thanks, I think this makes sense.
      – Boris Grunwald
      8 hours ago










    • And sorry, it was the smallest possible degree
      – Boris Grunwald
      8 hours ago
















    • Thanks, I think this makes sense.
      – Boris Grunwald
      8 hours ago










    • And sorry, it was the smallest possible degree
      – Boris Grunwald
      8 hours ago















    Thanks, I think this makes sense.
    – Boris Grunwald
    8 hours ago




    Thanks, I think this makes sense.
    – Boris Grunwald
    8 hours ago












    And sorry, it was the smallest possible degree
    – Boris Grunwald
    8 hours ago




    And sorry, it was the smallest possible degree
    – Boris Grunwald
    8 hours ago










    up vote
    3
    down vote













    Polynomial with only real coefficients is
    $$( z-10) , ( z-2i+8),(z+2 i+8) , ( z-i-3) , ( z+i-3)\=
    z^5-118 z^3-68 z^2+3160 z-6800
    $$






    share|cite|improve this answer
























      up vote
      3
      down vote













      Polynomial with only real coefficients is
      $$( z-10) , ( z-2i+8),(z+2 i+8) , ( z-i-3) , ( z+i-3)\=
      z^5-118 z^3-68 z^2+3160 z-6800
      $$






      share|cite|improve this answer






















        up vote
        3
        down vote










        up vote
        3
        down vote









        Polynomial with only real coefficients is
        $$( z-10) , ( z-2i+8),(z+2 i+8) , ( z-i-3) , ( z+i-3)\=
        z^5-118 z^3-68 z^2+3160 z-6800
        $$






        share|cite|improve this answer












        Polynomial with only real coefficients is
        $$( z-10) , ( z-2i+8),(z+2 i+8) , ( z-i-3) , ( z+i-3)\=
        z^5-118 z^3-68 z^2+3160 z-6800
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 8 hours ago









        Aleksas Domarkas

        5154




        5154



























             

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