Finding real $a$, $b$, $c$ such that $x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$ has $1+i$ as a zero, and one...












1












$begingroup$



Find $a, b, c in mathbb{R}$, if one zero of $p$ is $1+i$ and $p$ has one negative integer zero with multiplicity $2$, where:
$$p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$$




I know that one zero of $p$ is $1-i$, so I can reduce it one degree down after dividing with $x-2$, but after that I'm stuck.



I tried using Vieta's formulas, but with no result.






polynomials complex-numbers roots







share|cite|improve this question











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  • $begingroup$
    You're mistaken to use $(x-2)$ as a factor. Instead, it should be $(x-q)^2$ for some unknown $0 < q in mathbb{Z}$.
    $endgroup$
    – David G. Stork
    Jan 23 at 20:10








  • 1




    $begingroup$
    You have to divide by $(x-1-i)(x-1+i)$, not $x-2$.
    $endgroup$
    – orion
    Jan 23 at 20:11








  • 1




    $begingroup$
    @onion: No. $p(x) = (x - (1+i))(x-q)^2 r(x)$ where $r(x)$ is a quadratic. "Multiplicity $2$" means you have two roots that are the same... not that the value has anything to do with $2$. Your revision: again NO. The multiplicity-2 root is different from the root at $1 + i$. Please read carefully.
    $endgroup$
    – David G. Stork
    Jan 23 at 20:14












  • $begingroup$
    Oh right, i missed that. I'll try fixing it now.
    $endgroup$
    – user626177
    Jan 23 at 20:16










  • $begingroup$
    @DavidG.Stork You can always divide out both known complex roots to reduce the degree by two! That's before discussing the rest of the roots (where the multiplicity information is used).
    $endgroup$
    – orion
    Jan 23 at 20:38
















1












$begingroup$



Find $a, b, c in mathbb{R}$, if one zero of $p$ is $1+i$ and $p$ has one negative integer zero with multiplicity $2$, where:
$$p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$$




I know that one zero of $p$ is $1-i$, so I can reduce it one degree down after dividing with $x-2$, but after that I'm stuck.



I tried using Vieta's formulas, but with no result.






polynomials complex-numbers roots







share|cite|improve this question











$endgroup$












  • $begingroup$
    You're mistaken to use $(x-2)$ as a factor. Instead, it should be $(x-q)^2$ for some unknown $0 < q in mathbb{Z}$.
    $endgroup$
    – David G. Stork
    Jan 23 at 20:10








  • 1




    $begingroup$
    You have to divide by $(x-1-i)(x-1+i)$, not $x-2$.
    $endgroup$
    – orion
    Jan 23 at 20:11








  • 1




    $begingroup$
    @onion: No. $p(x) = (x - (1+i))(x-q)^2 r(x)$ where $r(x)$ is a quadratic. "Multiplicity $2$" means you have two roots that are the same... not that the value has anything to do with $2$. Your revision: again NO. The multiplicity-2 root is different from the root at $1 + i$. Please read carefully.
    $endgroup$
    – David G. Stork
    Jan 23 at 20:14












  • $begingroup$
    Oh right, i missed that. I'll try fixing it now.
    $endgroup$
    – user626177
    Jan 23 at 20:16










  • $begingroup$
    @DavidG.Stork You can always divide out both known complex roots to reduce the degree by two! That's before discussing the rest of the roots (where the multiplicity information is used).
    $endgroup$
    – orion
    Jan 23 at 20:38














1












1








1





$begingroup$



Find $a, b, c in mathbb{R}$, if one zero of $p$ is $1+i$ and $p$ has one negative integer zero with multiplicity $2$, where:
$$p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$$




I know that one zero of $p$ is $1-i$, so I can reduce it one degree down after dividing with $x-2$, but after that I'm stuck.



I tried using Vieta's formulas, but with no result.






polynomials complex-numbers roots







share|cite|improve this question











$endgroup$





Find $a, b, c in mathbb{R}$, if one zero of $p$ is $1+i$ and $p$ has one negative integer zero with multiplicity $2$, where:
$$p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$$




I know that one zero of $p$ is $1-i$, so I can reduce it one degree down after dividing with $x-2$, but after that I'm stuck.



I tried using Vieta's formulas, but with no result.






polynomials complex-numbers roots




polynomials complex-numbers roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 20:42









Michael Rozenberg

106k1894198




106k1894198










asked Jan 23 at 20:01







user626177



















  • $begingroup$
    You're mistaken to use $(x-2)$ as a factor. Instead, it should be $(x-q)^2$ for some unknown $0 < q in mathbb{Z}$.
    $endgroup$
    – David G. Stork
    Jan 23 at 20:10








  • 1




    $begingroup$
    You have to divide by $(x-1-i)(x-1+i)$, not $x-2$.
    $endgroup$
    – orion
    Jan 23 at 20:11








  • 1




    $begingroup$
    @onion: No. $p(x) = (x - (1+i))(x-q)^2 r(x)$ where $r(x)$ is a quadratic. "Multiplicity $2$" means you have two roots that are the same... not that the value has anything to do with $2$. Your revision: again NO. The multiplicity-2 root is different from the root at $1 + i$. Please read carefully.
    $endgroup$
    – David G. Stork
    Jan 23 at 20:14












  • $begingroup$
    Oh right, i missed that. I'll try fixing it now.
    $endgroup$
    – user626177
    Jan 23 at 20:16










  • $begingroup$
    @DavidG.Stork You can always divide out both known complex roots to reduce the degree by two! That's before discussing the rest of the roots (where the multiplicity information is used).
    $endgroup$
    – orion
    Jan 23 at 20:38


















  • $begingroup$
    You're mistaken to use $(x-2)$ as a factor. Instead, it should be $(x-q)^2$ for some unknown $0 < q in mathbb{Z}$.
    $endgroup$
    – David G. Stork
    Jan 23 at 20:10








  • 1




    $begingroup$
    You have to divide by $(x-1-i)(x-1+i)$, not $x-2$.
    $endgroup$
    – orion
    Jan 23 at 20:11








  • 1




    $begingroup$
    @onion: No. $p(x) = (x - (1+i))(x-q)^2 r(x)$ where $r(x)$ is a quadratic. "Multiplicity $2$" means you have two roots that are the same... not that the value has anything to do with $2$. Your revision: again NO. The multiplicity-2 root is different from the root at $1 + i$. Please read carefully.
    $endgroup$
    – David G. Stork
    Jan 23 at 20:14












  • $begingroup$
    Oh right, i missed that. I'll try fixing it now.
    $endgroup$
    – user626177
    Jan 23 at 20:16










  • $begingroup$
    @DavidG.Stork You can always divide out both known complex roots to reduce the degree by two! That's before discussing the rest of the roots (where the multiplicity information is used).
    $endgroup$
    – orion
    Jan 23 at 20:38
















$begingroup$
You're mistaken to use $(x-2)$ as a factor. Instead, it should be $(x-q)^2$ for some unknown $0 < q in mathbb{Z}$.
$endgroup$
– David G. Stork
Jan 23 at 20:10






$begingroup$
You're mistaken to use $(x-2)$ as a factor. Instead, it should be $(x-q)^2$ for some unknown $0 < q in mathbb{Z}$.
$endgroup$
– David G. Stork
Jan 23 at 20:10






1




1




$begingroup$
You have to divide by $(x-1-i)(x-1+i)$, not $x-2$.
$endgroup$
– orion
Jan 23 at 20:11






$begingroup$
You have to divide by $(x-1-i)(x-1+i)$, not $x-2$.
$endgroup$
– orion
Jan 23 at 20:11






1




1




$begingroup$
@onion: No. $p(x) = (x - (1+i))(x-q)^2 r(x)$ where $r(x)$ is a quadratic. "Multiplicity $2$" means you have two roots that are the same... not that the value has anything to do with $2$. Your revision: again NO. The multiplicity-2 root is different from the root at $1 + i$. Please read carefully.
$endgroup$
– David G. Stork
Jan 23 at 20:14






$begingroup$
@onion: No. $p(x) = (x - (1+i))(x-q)^2 r(x)$ where $r(x)$ is a quadratic. "Multiplicity $2$" means you have two roots that are the same... not that the value has anything to do with $2$. Your revision: again NO. The multiplicity-2 root is different from the root at $1 + i$. Please read carefully.
$endgroup$
– David G. Stork
Jan 23 at 20:14














$begingroup$
Oh right, i missed that. I'll try fixing it now.
$endgroup$
– user626177
Jan 23 at 20:16




$begingroup$
Oh right, i missed that. I'll try fixing it now.
$endgroup$
– user626177
Jan 23 at 20:16












$begingroup$
@DavidG.Stork You can always divide out both known complex roots to reduce the degree by two! That's before discussing the rest of the roots (where the multiplicity information is used).
$endgroup$
– orion
Jan 23 at 20:38




$begingroup$
@DavidG.Stork You can always divide out both known complex roots to reduce the degree by two! That's before discussing the rest of the roots (where the multiplicity information is used).
$endgroup$
– orion
Jan 23 at 20:38










3 Answers
3






active

oldest

votes


















0












$begingroup$

Since all coefficients are reals, $p$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2+1=x^2-2x+2.$$



We have $$p(x)=x^5-2x^4+ax^3+bx^2-2x+c=$$
$$=x^5-2x^4+2x^3+(a-2)x^3-2(a-2)x^2+2(a-2)x+(b+2a-4)x^2-(2a-2)x+c=$$
$$=(x^2-2x+2)(x^3+(a-2)x)+(b+2a-4)x^2-(2a-2)x+c.$$
We need $$b+2a-4=k,$$ $$2a-2=2k$$ and $$c=2k,$$ which gives
$$b=3-a,$$ $$c=2a-2$$ and
$$p(x)=(x^2-2x+2)((x^3-2x-1+a(x+1)).$$
Now, let $x^3-2x-1+a(x+1)=q(x).$



Thus, $q$ and $q'$ have the same negative integer root.



$$q'(x)=3x^2-2+a,$$ which gives $$a=2-3x^2.$$



Id est, for the same root we obtain the following equation.
$$x^3-2x-1+(2-3x^2)(x+1)=0$$ or
$$2x^3+3x^2-1=0$$ or
$$2x^3+4x^2+2x-x^2-2x-1=0$$ or
$$(x+1)^2(2x-1)=0,$$
which gives $x=-1$ and $a=-1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have root of derivative of $q$ in terms of $a$, so now I plug that in in $q$ ? Or you've done it somehow faster .. ?
    $endgroup$
    – user626177
    Jan 23 at 20:35










  • $begingroup$
    Just, $3x^2-2+a=0$, which gives $2x^3+3x^2-1=0$ and from here $x=-1$.
    $endgroup$
    – Michael Rozenberg
    Jan 23 at 20:36












  • $begingroup$
    Can you please explain that, I don't follow how from $3x^2 -2 +a =0$ you derived $2x^3 +3x^2 -1=0$ ?
    $endgroup$
    – user626177
    Jan 23 at 20:44






  • 1




    $begingroup$
    @someone I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 23 at 20:50



















1












$begingroup$

$$x_1=1+i$$
$$x_2=1-i$$
$$x_3=x_4 = m<0$$
and $$x_5=k$$ then we have $$x_1+x_2+x_3+x_4+x_5=2$$



$$2+2m+k=2implies boxed{k=-2m}$$
and $$x_1x_2x_3x_4x_5({1over x_1}+{1over x_2}+{1over x_3}+{1over x_4}+{1over x_5}) =-2$$



so $$boxed{2m^2k(1+{2over m}+{1over k})=-2}implies m^2(2m+3)=-1$$



Since $m^2mid -1$ we have $m^2=1implies m=-1$ and $k=2$.



The rest should be easy now.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @DavidG.Stork If we know $p$ has one complex zero $z$, then $z$ conjugate is also zero, if $p$ is with real coefficients (if you are talking about that)
    $endgroup$
    – user626177
    Jan 23 at 20:27





















0












$begingroup$

If $ 1 + i $ is the root of the polynomial
$$
p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
$$
and $a,b,cinmathbb{R}$ then $1-i$ is root of $p(x)$. Soon $q(x)=(x-1+i)cdot(x-1-i)=(x^2-2x+2)$ divide $p(x)$ and
begin{align}
x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
=&
(x^2-2x+2)big( x^3 -(2r+s)x^2+(r^2+2rs)x-r^2sbig)
\
=&x^5
\
&hspace{0.5cm} -2x^4-(2r+s)x^4
\
&hspace{1.0cm} +2x^3+2(2r+s)x^3+(r^2+2rs)x^3
\
&hspace{1.5cm} -2(2r+s)x^2-2(r^2+2rs)x^2-r^2sx^2
\
&hspace{2.0cm} +2(r^2+2rs)x+2r^2sx
\
&hspace{2.5cm} +2(r^2+2rs)x+2r^2sx
\
&hspace{3.0cm} -2r^2s
end{align}
Note that
begin{align}
-2-(2r+s)=&-2quad implies quad 2r+s=0\
+2+2(2r+s)+(r^2+2rs)=&a\
-2(2r+s)-2(r^2+2rs)-r^2s=&b\
+2(r^2+2rs)+2r^2s=&-2 quad implies quad 2r^2+4rs+2r^2s=-2\
-2r^2s=&c\
end{align}






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Since all coefficients are reals, $p$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2+1=x^2-2x+2.$$



    We have $$p(x)=x^5-2x^4+ax^3+bx^2-2x+c=$$
    $$=x^5-2x^4+2x^3+(a-2)x^3-2(a-2)x^2+2(a-2)x+(b+2a-4)x^2-(2a-2)x+c=$$
    $$=(x^2-2x+2)(x^3+(a-2)x)+(b+2a-4)x^2-(2a-2)x+c.$$
    We need $$b+2a-4=k,$$ $$2a-2=2k$$ and $$c=2k,$$ which gives
    $$b=3-a,$$ $$c=2a-2$$ and
    $$p(x)=(x^2-2x+2)((x^3-2x-1+a(x+1)).$$
    Now, let $x^3-2x-1+a(x+1)=q(x).$



    Thus, $q$ and $q'$ have the same negative integer root.



    $$q'(x)=3x^2-2+a,$$ which gives $$a=2-3x^2.$$



    Id est, for the same root we obtain the following equation.
    $$x^3-2x-1+(2-3x^2)(x+1)=0$$ or
    $$2x^3+3x^2-1=0$$ or
    $$2x^3+4x^2+2x-x^2-2x-1=0$$ or
    $$(x+1)^2(2x-1)=0,$$
    which gives $x=-1$ and $a=-1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have root of derivative of $q$ in terms of $a$, so now I plug that in in $q$ ? Or you've done it somehow faster .. ?
      $endgroup$
      – user626177
      Jan 23 at 20:35










    • $begingroup$
      Just, $3x^2-2+a=0$, which gives $2x^3+3x^2-1=0$ and from here $x=-1$.
      $endgroup$
      – Michael Rozenberg
      Jan 23 at 20:36












    • $begingroup$
      Can you please explain that, I don't follow how from $3x^2 -2 +a =0$ you derived $2x^3 +3x^2 -1=0$ ?
      $endgroup$
      – user626177
      Jan 23 at 20:44






    • 1




      $begingroup$
      @someone I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Jan 23 at 20:50
















    0












    $begingroup$

    Since all coefficients are reals, $p$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2+1=x^2-2x+2.$$



    We have $$p(x)=x^5-2x^4+ax^3+bx^2-2x+c=$$
    $$=x^5-2x^4+2x^3+(a-2)x^3-2(a-2)x^2+2(a-2)x+(b+2a-4)x^2-(2a-2)x+c=$$
    $$=(x^2-2x+2)(x^3+(a-2)x)+(b+2a-4)x^2-(2a-2)x+c.$$
    We need $$b+2a-4=k,$$ $$2a-2=2k$$ and $$c=2k,$$ which gives
    $$b=3-a,$$ $$c=2a-2$$ and
    $$p(x)=(x^2-2x+2)((x^3-2x-1+a(x+1)).$$
    Now, let $x^3-2x-1+a(x+1)=q(x).$



    Thus, $q$ and $q'$ have the same negative integer root.



    $$q'(x)=3x^2-2+a,$$ which gives $$a=2-3x^2.$$



    Id est, for the same root we obtain the following equation.
    $$x^3-2x-1+(2-3x^2)(x+1)=0$$ or
    $$2x^3+3x^2-1=0$$ or
    $$2x^3+4x^2+2x-x^2-2x-1=0$$ or
    $$(x+1)^2(2x-1)=0,$$
    which gives $x=-1$ and $a=-1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have root of derivative of $q$ in terms of $a$, so now I plug that in in $q$ ? Or you've done it somehow faster .. ?
      $endgroup$
      – user626177
      Jan 23 at 20:35










    • $begingroup$
      Just, $3x^2-2+a=0$, which gives $2x^3+3x^2-1=0$ and from here $x=-1$.
      $endgroup$
      – Michael Rozenberg
      Jan 23 at 20:36












    • $begingroup$
      Can you please explain that, I don't follow how from $3x^2 -2 +a =0$ you derived $2x^3 +3x^2 -1=0$ ?
      $endgroup$
      – user626177
      Jan 23 at 20:44






    • 1




      $begingroup$
      @someone I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Jan 23 at 20:50














    0












    0








    0





    $begingroup$

    Since all coefficients are reals, $p$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2+1=x^2-2x+2.$$



    We have $$p(x)=x^5-2x^4+ax^3+bx^2-2x+c=$$
    $$=x^5-2x^4+2x^3+(a-2)x^3-2(a-2)x^2+2(a-2)x+(b+2a-4)x^2-(2a-2)x+c=$$
    $$=(x^2-2x+2)(x^3+(a-2)x)+(b+2a-4)x^2-(2a-2)x+c.$$
    We need $$b+2a-4=k,$$ $$2a-2=2k$$ and $$c=2k,$$ which gives
    $$b=3-a,$$ $$c=2a-2$$ and
    $$p(x)=(x^2-2x+2)((x^3-2x-1+a(x+1)).$$
    Now, let $x^3-2x-1+a(x+1)=q(x).$



    Thus, $q$ and $q'$ have the same negative integer root.



    $$q'(x)=3x^2-2+a,$$ which gives $$a=2-3x^2.$$



    Id est, for the same root we obtain the following equation.
    $$x^3-2x-1+(2-3x^2)(x+1)=0$$ or
    $$2x^3+3x^2-1=0$$ or
    $$2x^3+4x^2+2x-x^2-2x-1=0$$ or
    $$(x+1)^2(2x-1)=0,$$
    which gives $x=-1$ and $a=-1$.






    share|cite|improve this answer











    $endgroup$



    Since all coefficients are reals, $p$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2+1=x^2-2x+2.$$



    We have $$p(x)=x^5-2x^4+ax^3+bx^2-2x+c=$$
    $$=x^5-2x^4+2x^3+(a-2)x^3-2(a-2)x^2+2(a-2)x+(b+2a-4)x^2-(2a-2)x+c=$$
    $$=(x^2-2x+2)(x^3+(a-2)x)+(b+2a-4)x^2-(2a-2)x+c.$$
    We need $$b+2a-4=k,$$ $$2a-2=2k$$ and $$c=2k,$$ which gives
    $$b=3-a,$$ $$c=2a-2$$ and
    $$p(x)=(x^2-2x+2)((x^3-2x-1+a(x+1)).$$
    Now, let $x^3-2x-1+a(x+1)=q(x).$



    Thus, $q$ and $q'$ have the same negative integer root.



    $$q'(x)=3x^2-2+a,$$ which gives $$a=2-3x^2.$$



    Id est, for the same root we obtain the following equation.
    $$x^3-2x-1+(2-3x^2)(x+1)=0$$ or
    $$2x^3+3x^2-1=0$$ or
    $$2x^3+4x^2+2x-x^2-2x-1=0$$ or
    $$(x+1)^2(2x-1)=0,$$
    which gives $x=-1$ and $a=-1$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 23 at 20:49

























    answered Jan 23 at 20:24









    Michael RozenbergMichael Rozenberg

    106k1894198




    106k1894198












    • $begingroup$
      I have root of derivative of $q$ in terms of $a$, so now I plug that in in $q$ ? Or you've done it somehow faster .. ?
      $endgroup$
      – user626177
      Jan 23 at 20:35










    • $begingroup$
      Just, $3x^2-2+a=0$, which gives $2x^3+3x^2-1=0$ and from here $x=-1$.
      $endgroup$
      – Michael Rozenberg
      Jan 23 at 20:36












    • $begingroup$
      Can you please explain that, I don't follow how from $3x^2 -2 +a =0$ you derived $2x^3 +3x^2 -1=0$ ?
      $endgroup$
      – user626177
      Jan 23 at 20:44






    • 1




      $begingroup$
      @someone I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Jan 23 at 20:50


















    • $begingroup$
      I have root of derivative of $q$ in terms of $a$, so now I plug that in in $q$ ? Or you've done it somehow faster .. ?
      $endgroup$
      – user626177
      Jan 23 at 20:35










    • $begingroup$
      Just, $3x^2-2+a=0$, which gives $2x^3+3x^2-1=0$ and from here $x=-1$.
      $endgroup$
      – Michael Rozenberg
      Jan 23 at 20:36












    • $begingroup$
      Can you please explain that, I don't follow how from $3x^2 -2 +a =0$ you derived $2x^3 +3x^2 -1=0$ ?
      $endgroup$
      – user626177
      Jan 23 at 20:44






    • 1




      $begingroup$
      @someone I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Jan 23 at 20:50
















    $begingroup$
    I have root of derivative of $q$ in terms of $a$, so now I plug that in in $q$ ? Or you've done it somehow faster .. ?
    $endgroup$
    – user626177
    Jan 23 at 20:35




    $begingroup$
    I have root of derivative of $q$ in terms of $a$, so now I plug that in in $q$ ? Or you've done it somehow faster .. ?
    $endgroup$
    – user626177
    Jan 23 at 20:35












    $begingroup$
    Just, $3x^2-2+a=0$, which gives $2x^3+3x^2-1=0$ and from here $x=-1$.
    $endgroup$
    – Michael Rozenberg
    Jan 23 at 20:36






    $begingroup$
    Just, $3x^2-2+a=0$, which gives $2x^3+3x^2-1=0$ and from here $x=-1$.
    $endgroup$
    – Michael Rozenberg
    Jan 23 at 20:36














    $begingroup$
    Can you please explain that, I don't follow how from $3x^2 -2 +a =0$ you derived $2x^3 +3x^2 -1=0$ ?
    $endgroup$
    – user626177
    Jan 23 at 20:44




    $begingroup$
    Can you please explain that, I don't follow how from $3x^2 -2 +a =0$ you derived $2x^3 +3x^2 -1=0$ ?
    $endgroup$
    – user626177
    Jan 23 at 20:44




    1




    1




    $begingroup$
    @someone I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 23 at 20:50




    $begingroup$
    @someone I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 23 at 20:50











    1












    $begingroup$

    $$x_1=1+i$$
    $$x_2=1-i$$
    $$x_3=x_4 = m<0$$
    and $$x_5=k$$ then we have $$x_1+x_2+x_3+x_4+x_5=2$$



    $$2+2m+k=2implies boxed{k=-2m}$$
    and $$x_1x_2x_3x_4x_5({1over x_1}+{1over x_2}+{1over x_3}+{1over x_4}+{1over x_5}) =-2$$



    so $$boxed{2m^2k(1+{2over m}+{1over k})=-2}implies m^2(2m+3)=-1$$



    Since $m^2mid -1$ we have $m^2=1implies m=-1$ and $k=2$.



    The rest should be easy now.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @DavidG.Stork If we know $p$ has one complex zero $z$, then $z$ conjugate is also zero, if $p$ is with real coefficients (if you are talking about that)
      $endgroup$
      – user626177
      Jan 23 at 20:27


















    1












    $begingroup$

    $$x_1=1+i$$
    $$x_2=1-i$$
    $$x_3=x_4 = m<0$$
    and $$x_5=k$$ then we have $$x_1+x_2+x_3+x_4+x_5=2$$



    $$2+2m+k=2implies boxed{k=-2m}$$
    and $$x_1x_2x_3x_4x_5({1over x_1}+{1over x_2}+{1over x_3}+{1over x_4}+{1over x_5}) =-2$$



    so $$boxed{2m^2k(1+{2over m}+{1over k})=-2}implies m^2(2m+3)=-1$$



    Since $m^2mid -1$ we have $m^2=1implies m=-1$ and $k=2$.



    The rest should be easy now.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @DavidG.Stork If we know $p$ has one complex zero $z$, then $z$ conjugate is also zero, if $p$ is with real coefficients (if you are talking about that)
      $endgroup$
      – user626177
      Jan 23 at 20:27
















    1












    1








    1





    $begingroup$

    $$x_1=1+i$$
    $$x_2=1-i$$
    $$x_3=x_4 = m<0$$
    and $$x_5=k$$ then we have $$x_1+x_2+x_3+x_4+x_5=2$$



    $$2+2m+k=2implies boxed{k=-2m}$$
    and $$x_1x_2x_3x_4x_5({1over x_1}+{1over x_2}+{1over x_3}+{1over x_4}+{1over x_5}) =-2$$



    so $$boxed{2m^2k(1+{2over m}+{1over k})=-2}implies m^2(2m+3)=-1$$



    Since $m^2mid -1$ we have $m^2=1implies m=-1$ and $k=2$.



    The rest should be easy now.






    share|cite|improve this answer











    $endgroup$



    $$x_1=1+i$$
    $$x_2=1-i$$
    $$x_3=x_4 = m<0$$
    and $$x_5=k$$ then we have $$x_1+x_2+x_3+x_4+x_5=2$$



    $$2+2m+k=2implies boxed{k=-2m}$$
    and $$x_1x_2x_3x_4x_5({1over x_1}+{1over x_2}+{1over x_3}+{1over x_4}+{1over x_5}) =-2$$



    so $$boxed{2m^2k(1+{2over m}+{1over k})=-2}implies m^2(2m+3)=-1$$



    Since $m^2mid -1$ we have $m^2=1implies m=-1$ and $k=2$.



    The rest should be easy now.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 23 at 20:56

























    answered Jan 23 at 20:24









    greedoidgreedoid

    45.9k1160116




    45.9k1160116












    • $begingroup$
      @DavidG.Stork If we know $p$ has one complex zero $z$, then $z$ conjugate is also zero, if $p$ is with real coefficients (if you are talking about that)
      $endgroup$
      – user626177
      Jan 23 at 20:27




















    • $begingroup$
      @DavidG.Stork If we know $p$ has one complex zero $z$, then $z$ conjugate is also zero, if $p$ is with real coefficients (if you are talking about that)
      $endgroup$
      – user626177
      Jan 23 at 20:27


















    $begingroup$
    @DavidG.Stork If we know $p$ has one complex zero $z$, then $z$ conjugate is also zero, if $p$ is with real coefficients (if you are talking about that)
    $endgroup$
    – user626177
    Jan 23 at 20:27






    $begingroup$
    @DavidG.Stork If we know $p$ has one complex zero $z$, then $z$ conjugate is also zero, if $p$ is with real coefficients (if you are talking about that)
    $endgroup$
    – user626177
    Jan 23 at 20:27













    0












    $begingroup$

    If $ 1 + i $ is the root of the polynomial
    $$
    p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
    $$
    and $a,b,cinmathbb{R}$ then $1-i$ is root of $p(x)$. Soon $q(x)=(x-1+i)cdot(x-1-i)=(x^2-2x+2)$ divide $p(x)$ and
    begin{align}
    x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
    =&
    (x^2-2x+2)big( x^3 -(2r+s)x^2+(r^2+2rs)x-r^2sbig)
    \
    =&x^5
    \
    &hspace{0.5cm} -2x^4-(2r+s)x^4
    \
    &hspace{1.0cm} +2x^3+2(2r+s)x^3+(r^2+2rs)x^3
    \
    &hspace{1.5cm} -2(2r+s)x^2-2(r^2+2rs)x^2-r^2sx^2
    \
    &hspace{2.0cm} +2(r^2+2rs)x+2r^2sx
    \
    &hspace{2.5cm} +2(r^2+2rs)x+2r^2sx
    \
    &hspace{3.0cm} -2r^2s
    end{align}
    Note that
    begin{align}
    -2-(2r+s)=&-2quad implies quad 2r+s=0\
    +2+2(2r+s)+(r^2+2rs)=&a\
    -2(2r+s)-2(r^2+2rs)-r^2s=&b\
    +2(r^2+2rs)+2r^2s=&-2 quad implies quad 2r^2+4rs+2r^2s=-2\
    -2r^2s=&c\
    end{align}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If $ 1 + i $ is the root of the polynomial
      $$
      p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
      $$
      and $a,b,cinmathbb{R}$ then $1-i$ is root of $p(x)$. Soon $q(x)=(x-1+i)cdot(x-1-i)=(x^2-2x+2)$ divide $p(x)$ and
      begin{align}
      x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
      =&
      (x^2-2x+2)big( x^3 -(2r+s)x^2+(r^2+2rs)x-r^2sbig)
      \
      =&x^5
      \
      &hspace{0.5cm} -2x^4-(2r+s)x^4
      \
      &hspace{1.0cm} +2x^3+2(2r+s)x^3+(r^2+2rs)x^3
      \
      &hspace{1.5cm} -2(2r+s)x^2-2(r^2+2rs)x^2-r^2sx^2
      \
      &hspace{2.0cm} +2(r^2+2rs)x+2r^2sx
      \
      &hspace{2.5cm} +2(r^2+2rs)x+2r^2sx
      \
      &hspace{3.0cm} -2r^2s
      end{align}
      Note that
      begin{align}
      -2-(2r+s)=&-2quad implies quad 2r+s=0\
      +2+2(2r+s)+(r^2+2rs)=&a\
      -2(2r+s)-2(r^2+2rs)-r^2s=&b\
      +2(r^2+2rs)+2r^2s=&-2 quad implies quad 2r^2+4rs+2r^2s=-2\
      -2r^2s=&c\
      end{align}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If $ 1 + i $ is the root of the polynomial
        $$
        p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
        $$
        and $a,b,cinmathbb{R}$ then $1-i$ is root of $p(x)$. Soon $q(x)=(x-1+i)cdot(x-1-i)=(x^2-2x+2)$ divide $p(x)$ and
        begin{align}
        x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
        =&
        (x^2-2x+2)big( x^3 -(2r+s)x^2+(r^2+2rs)x-r^2sbig)
        \
        =&x^5
        \
        &hspace{0.5cm} -2x^4-(2r+s)x^4
        \
        &hspace{1.0cm} +2x^3+2(2r+s)x^3+(r^2+2rs)x^3
        \
        &hspace{1.5cm} -2(2r+s)x^2-2(r^2+2rs)x^2-r^2sx^2
        \
        &hspace{2.0cm} +2(r^2+2rs)x+2r^2sx
        \
        &hspace{2.5cm} +2(r^2+2rs)x+2r^2sx
        \
        &hspace{3.0cm} -2r^2s
        end{align}
        Note that
        begin{align}
        -2-(2r+s)=&-2quad implies quad 2r+s=0\
        +2+2(2r+s)+(r^2+2rs)=&a\
        -2(2r+s)-2(r^2+2rs)-r^2s=&b\
        +2(r^2+2rs)+2r^2s=&-2 quad implies quad 2r^2+4rs+2r^2s=-2\
        -2r^2s=&c\
        end{align}






        share|cite|improve this answer









        $endgroup$



        If $ 1 + i $ is the root of the polynomial
        $$
        p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
        $$
        and $a,b,cinmathbb{R}$ then $1-i$ is root of $p(x)$. Soon $q(x)=(x-1+i)cdot(x-1-i)=(x^2-2x+2)$ divide $p(x)$ and
        begin{align}
        x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
        =&
        (x^2-2x+2)big( x^3 -(2r+s)x^2+(r^2+2rs)x-r^2sbig)
        \
        =&x^5
        \
        &hspace{0.5cm} -2x^4-(2r+s)x^4
        \
        &hspace{1.0cm} +2x^3+2(2r+s)x^3+(r^2+2rs)x^3
        \
        &hspace{1.5cm} -2(2r+s)x^2-2(r^2+2rs)x^2-r^2sx^2
        \
        &hspace{2.0cm} +2(r^2+2rs)x+2r^2sx
        \
        &hspace{2.5cm} +2(r^2+2rs)x+2r^2sx
        \
        &hspace{3.0cm} -2r^2s
        end{align}
        Note that
        begin{align}
        -2-(2r+s)=&-2quad implies quad 2r+s=0\
        +2+2(2r+s)+(r^2+2rs)=&a\
        -2(2r+s)-2(r^2+2rs)-r^2s=&b\
        +2(r^2+2rs)+2r^2s=&-2 quad implies quad 2r^2+4rs+2r^2s=-2\
        -2r^2s=&c\
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 23 at 20:54









        MathOverviewMathOverview

        8,93543164




        8,93543164






























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