Find $p$ if the remainder when $(x^2 + px + 13)$ divided by $(x-2)$ is twice the remainder when it is divided...












2












$begingroup$


As far as I understand, this should be a simultaneous equation, although I'm not really sure where to go beyond that. I made the remainder of $f(-2)=r$ and $f(2)=2r$, and substituted the respective values.



$x^2+px+13$



With $(x-2)$



$f(2)=(2)^2+p(2)+13$



$f(2)=17+2p=2r$



With $(x+2)$



$f(-2)=(-2)^2+p(-2)+13$



$f(-2)=17-2p=r$



I guess I could do a simultaneous here, and I got $17/6$ as my result, but I'm not sure if that's correct.






proof-verification polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    looks good to me
    $endgroup$
    – randomgirl
    Dec 17 '18 at 16:20
















2












$begingroup$


As far as I understand, this should be a simultaneous equation, although I'm not really sure where to go beyond that. I made the remainder of $f(-2)=r$ and $f(2)=2r$, and substituted the respective values.



$x^2+px+13$



With $(x-2)$



$f(2)=(2)^2+p(2)+13$



$f(2)=17+2p=2r$



With $(x+2)$



$f(-2)=(-2)^2+p(-2)+13$



$f(-2)=17-2p=r$



I guess I could do a simultaneous here, and I got $17/6$ as my result, but I'm not sure if that's correct.






proof-verification polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    looks good to me
    $endgroup$
    – randomgirl
    Dec 17 '18 at 16:20














2












2








2





$begingroup$


As far as I understand, this should be a simultaneous equation, although I'm not really sure where to go beyond that. I made the remainder of $f(-2)=r$ and $f(2)=2r$, and substituted the respective values.



$x^2+px+13$



With $(x-2)$



$f(2)=(2)^2+p(2)+13$



$f(2)=17+2p=2r$



With $(x+2)$



$f(-2)=(-2)^2+p(-2)+13$



$f(-2)=17-2p=r$



I guess I could do a simultaneous here, and I got $17/6$ as my result, but I'm not sure if that's correct.






proof-verification polynomials







share|cite|improve this question











$endgroup$




As far as I understand, this should be a simultaneous equation, although I'm not really sure where to go beyond that. I made the remainder of $f(-2)=r$ and $f(2)=2r$, and substituted the respective values.



$x^2+px+13$



With $(x-2)$



$f(2)=(2)^2+p(2)+13$



$f(2)=17+2p=2r$



With $(x+2)$



$f(-2)=(-2)^2+p(-2)+13$



$f(-2)=17-2p=r$



I guess I could do a simultaneous here, and I got $17/6$ as my result, but I'm not sure if that's correct.






proof-verification polynomials




proof-verification polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 16:08









Tianlalu

3,09421138




3,09421138










asked Dec 17 '18 at 16:04









Roo23Roo23

213




213








  • 1




    $begingroup$
    looks good to me
    $endgroup$
    – randomgirl
    Dec 17 '18 at 16:20














  • 1




    $begingroup$
    looks good to me
    $endgroup$
    – randomgirl
    Dec 17 '18 at 16:20








1




1




$begingroup$
looks good to me
$endgroup$
– randomgirl
Dec 17 '18 at 16:20




$begingroup$
looks good to me
$endgroup$
– randomgirl
Dec 17 '18 at 16:20










2 Answers
2






active

oldest

votes


















0












$begingroup$

You are making it too hard.



$f(2) = 17+2p,
f(-2) = 17-2p$.



So you want
$17+2p = 2(17-2p)
=34-4p$
so
$17=6p$
or
$p = 17/6$
as you got.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    HINT:



    $$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$



    Via polynomial long division.






    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%2f3044117%2ffind-p-if-the-remainder-when-x2-px-13-divided-by-x-2-is-twice-the%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      You are making it too hard.



      $f(2) = 17+2p,
      f(-2) = 17-2p$      .



      So you want
      $17+2p = 2(17-2p)
      =34-4p$
      so
      $17=6p$
      or
      $p = 17/6$
      as you got.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        You are making it too hard.



        $f(2) = 17+2p,
        f(-2) = 17-2p$        .



        So you want
        $17+2p = 2(17-2p)
        =34-4p$
        so
        $17=6p$
        or
        $p = 17/6$
        as you got.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          You are making it too hard.



          $f(2) = 17+2p,
          f(-2) = 17-2p$          .



          So you want
          $17+2p = 2(17-2p)
          =34-4p$
          so
          $17=6p$
          or
          $p = 17/6$
          as you got.






          share|cite|improve this answer









          $endgroup$



          You are making it too hard.



          $f(2) = 17+2p,
          f(-2) = 17-2p$          .



          So you want
          $17+2p = 2(17-2p)
          =34-4p$
          so
          $17=6p$
          or
          $p = 17/6$
          as you got.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 18:15









          marty cohenmarty cohen

          74.4k549129




          74.4k549129























              0












              $begingroup$

              HINT:



              $$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$



              Via polynomial long division.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                HINT:



                $$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$



                Via polynomial long division.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  HINT:



                  $$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$



                  Via polynomial long division.






                  share|cite|improve this answer









                  $endgroup$



                  HINT:



                  $$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$



                  Via polynomial long division.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 16:21









                  Rhys HughesRhys Hughes

                  7,0251630




                  7,0251630






























                      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%2f3044117%2ffind-p-if-the-remainder-when-x2-px-13-divided-by-x-2-is-twice-the%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

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                      Read more

                      Puebla de Zaragoza

                      Image
                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                      Read more

                      Musa

                      Image
                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                      Read more