Root(zero) of a polynomial












-1












$begingroup$


How is the Root of a polynomial defined? I found on Wikipedia's page written: "A root of a polynomial is a zero of the corresponding polynomial function." Or the URL:https://en.m.wikipedia.org/wiki/Zero_of_a_function
Now if I want to find the root of polynomial 9a-2b then that would be it's corresponding function?






polynomials







share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    How is the Root of a polynomial defined? I found on Wikipedia's page written: "A root of a polynomial is a zero of the corresponding polynomial function." Or the URL:https://en.m.wikipedia.org/wiki/Zero_of_a_function
    Now if I want to find the root of polynomial 9a-2b then that would be it's corresponding function?






    polynomials







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1





      $begingroup$


      How is the Root of a polynomial defined? I found on Wikipedia's page written: "A root of a polynomial is a zero of the corresponding polynomial function." Or the URL:https://en.m.wikipedia.org/wiki/Zero_of_a_function
      Now if I want to find the root of polynomial 9a-2b then that would be it's corresponding function?






      polynomials







      share|cite|improve this question









      $endgroup$




      How is the Root of a polynomial defined? I found on Wikipedia's page written: "A root of a polynomial is a zero of the corresponding polynomial function." Or the URL:https://en.m.wikipedia.org/wiki/Zero_of_a_function
      Now if I want to find the root of polynomial 9a-2b then that would be it's corresponding function?






      polynomials




      polynomials






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 30 '18 at 8:57







      user629353





























          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
            $endgroup$
            – user629353
            Dec 30 '18 at 9:02










          • $begingroup$
            It's the functions that maps the pair $(a,b)$ into $9a-2b$.
            $endgroup$
            – José Carlos Santos
            Dec 30 '18 at 9:05



















          1












          $begingroup$

          A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            $9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.



            If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.






            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%2f3056627%2frootzero-of-a-polynomial%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









              1












              $begingroup$

              If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
                $endgroup$
                – user629353
                Dec 30 '18 at 9:02










              • $begingroup$
                It's the functions that maps the pair $(a,b)$ into $9a-2b$.
                $endgroup$
                – José Carlos Santos
                Dec 30 '18 at 9:05
















              1












              $begingroup$

              If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
                $endgroup$
                – user629353
                Dec 30 '18 at 9:02










              • $begingroup$
                It's the functions that maps the pair $(a,b)$ into $9a-2b$.
                $endgroup$
                – José Carlos Santos
                Dec 30 '18 at 9:05














              1












              1








              1





              $begingroup$

              If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$






              share|cite|improve this answer









              $endgroup$



              If you are working over a field $F$, then the function will be$$begin{array}{ccc}Ftimes F&longrightarrow&F\(a,b)&mapsto&9a-2b.end{array}$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 30 '18 at 9:00









              José Carlos SantosJosé Carlos Santos

              164k22131234




              164k22131234












              • $begingroup$
                Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
                $endgroup$
                – user629353
                Dec 30 '18 at 9:02










              • $begingroup$
                It's the functions that maps the pair $(a,b)$ into $9a-2b$.
                $endgroup$
                – José Carlos Santos
                Dec 30 '18 at 9:05


















              • $begingroup$
                Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
                $endgroup$
                – user629353
                Dec 30 '18 at 9:02










              • $begingroup$
                It's the functions that maps the pair $(a,b)$ into $9a-2b$.
                $endgroup$
                – José Carlos Santos
                Dec 30 '18 at 9:05
















              $begingroup$
              Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
              $endgroup$
              – user629353
              Dec 30 '18 at 9:02




              $begingroup$
              Sorry, but I am unfamiliar with abstract algebra topics, i.e fields, etc. Kindly tell in layman language
              $endgroup$
              – user629353
              Dec 30 '18 at 9:02












              $begingroup$
              It's the functions that maps the pair $(a,b)$ into $9a-2b$.
              $endgroup$
              – José Carlos Santos
              Dec 30 '18 at 9:05




              $begingroup$
              It's the functions that maps the pair $(a,b)$ into $9a-2b$.
              $endgroup$
              – José Carlos Santos
              Dec 30 '18 at 9:05











              1












              $begingroup$

              A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.






                  share|cite|improve this answer









                  $endgroup$



                  A root of a polynomial is the value (or values) given to its variable (or variables) that results in the polynomial evaluating to zero. For example, the root of a polynomial $x+2=0$ is $x=-2$. Likewise, roots of the polynomial $x^2-1=0$ are $x=pm 1$. In your case, the polynomial has two variables. But we have only one equation $9a-2b=0$. Therefore, the roots of this polynomial are any pair of numbers $a$ and $9a/2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 30 '18 at 9:05









                  Amey JoshiAmey Joshi

                  604310




                  604310























                      0












                      $begingroup$

                      $9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.



                      If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        $9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.



                        If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          $9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.



                          If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.






                          share|cite|improve this answer









                          $endgroup$



                          $9a-2b$ is not a polynomial without any more context. It is a linear formula in two unknowns, so we could indeed see it as a polynomial in the two variables $a,b$, say $p(a,b) = 9a-2b$ (which also defines a function). Over the reals, or the rational numbers, this has many zeroes, e.g. $(1,4frac{1}{2})$ is a root of the polynomial in $mathbb{Q}^2$, because if we fill in the values in $p$ we get $0$.



                          If $b$ were some known number or parameter, then we could also see it as a polynomial in $a$: $q(a) = 9a-2b$, and then it has only one root, if $b$ is in some field: $a = frac{2b}{9}$, which lies in the same field as $b$ provided $9$ has an inverse (it probably has, as I think you're working over real, complex or rational numbers?). In this case $b$ is not seen as an "unknown" or variable that can be substituted.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 30 '18 at 9:29









                          Henno BrandsmaHenno Brandsma

                          111k348118




                          111k348118






























                              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%2f3056627%2frootzero-of-a-polynomial%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

                              Bressuire

                              Image
                              .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                              Read more

                              Cabo Verde

                              Image
                                Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                              Read more

                              Gyllenstierna

                              Image
                              .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                              Read more