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











$endgroup$












  • $begingroup$
    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












$begingroup$



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}






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084994%2ffinding-real-a-b-c-such-that-x5-2x4-ax3-bx2-2x-c-has-1%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown
























    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084994%2ffinding-real-a-b-c-such-that-x5-2x4-ax3-bx2-2x-c-has-1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me
            Read more