Why does this Polynomial have two valid factored forms?












1












$begingroup$


When factoring $3 + 4a - 7a^2$ why are there two possible factors? $(-7a -3)(a - 1)$ and $(3 + 7a)(1 - a)$. What is the principle? I assume that the two factors are inverse in some way.










share|cite|improve this question











$endgroup$












  • $begingroup$
    You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
    $endgroup$
    – lulu
    Dec 22 '18 at 16:25








  • 1




    $begingroup$
    If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:26






  • 1




    $begingroup$
    Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
    $endgroup$
    – badjohn
    Dec 22 '18 at 16:33


















1












$begingroup$


When factoring $3 + 4a - 7a^2$ why are there two possible factors? $(-7a -3)(a - 1)$ and $(3 + 7a)(1 - a)$. What is the principle? I assume that the two factors are inverse in some way.










share|cite|improve this question











$endgroup$












  • $begingroup$
    You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
    $endgroup$
    – lulu
    Dec 22 '18 at 16:25








  • 1




    $begingroup$
    If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:26






  • 1




    $begingroup$
    Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
    $endgroup$
    – badjohn
    Dec 22 '18 at 16:33
















1












1








1





$begingroup$


When factoring $3 + 4a - 7a^2$ why are there two possible factors? $(-7a -3)(a - 1)$ and $(3 + 7a)(1 - a)$. What is the principle? I assume that the two factors are inverse in some way.










share|cite|improve this question











$endgroup$




When factoring $3 + 4a - 7a^2$ why are there two possible factors? $(-7a -3)(a - 1)$ and $(3 + 7a)(1 - a)$. What is the principle? I assume that the two factors are inverse in some way.







polynomials factoring






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 16:29







patawa91

















asked Dec 22 '18 at 16:24









patawa91patawa91

125




125












  • $begingroup$
    You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
    $endgroup$
    – lulu
    Dec 22 '18 at 16:25








  • 1




    $begingroup$
    If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:26






  • 1




    $begingroup$
    Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
    $endgroup$
    – badjohn
    Dec 22 '18 at 16:33




















  • $begingroup$
    You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
    $endgroup$
    – lulu
    Dec 22 '18 at 16:25








  • 1




    $begingroup$
    If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:26






  • 1




    $begingroup$
    Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
    $endgroup$
    – badjohn
    Dec 22 '18 at 16:33


















$begingroup$
You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
$endgroup$
– lulu
Dec 22 '18 at 16:25






$begingroup$
You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
$endgroup$
– lulu
Dec 22 '18 at 16:25






1




1




$begingroup$
If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
$endgroup$
– lulu
Dec 22 '18 at 16:26




$begingroup$
If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
$endgroup$
– lulu
Dec 22 '18 at 16:26




1




1




$begingroup$
Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
$endgroup$
– badjohn
Dec 22 '18 at 16:33






$begingroup$
Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
$endgroup$
– badjohn
Dec 22 '18 at 16:33












2 Answers
2






active

oldest

votes


















1












$begingroup$

(-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).



Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.



    Ex 1.



    -7a-3 = 0



    a = -3/7



    Ex 2.



    3+7a=0

    a = -3/7






    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%2f3049610%2fwhy-does-this-polynomial-have-two-valid-factored-forms%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









      1












      $begingroup$

      (-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).



      Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        (-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).



        Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          (-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).



          Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.






          share|cite|improve this answer











          $endgroup$



          (-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).



          Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 22 '18 at 16:35

























          answered Dec 22 '18 at 16:28









          MickMick

          11.9k21641




          11.9k21641























              0












              $begingroup$

              The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.



              Ex 1.



              -7a-3 = 0



              a = -3/7



              Ex 2.



              3+7a=0

              a = -3/7






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.



                Ex 1.



                -7a-3 = 0



                a = -3/7



                Ex 2.



                3+7a=0

                a = -3/7






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.



                  Ex 1.



                  -7a-3 = 0



                  a = -3/7



                  Ex 2.



                  3+7a=0

                  a = -3/7






                  share|cite|improve this answer









                  $endgroup$



                  The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.



                  Ex 1.



                  -7a-3 = 0



                  a = -3/7



                  Ex 2.



                  3+7a=0

                  a = -3/7







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 22 '18 at 16:35









                  muhe31muhe31

                  85




                  85






























                      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%2f3049610%2fwhy-does-this-polynomial-have-two-valid-factored-forms%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

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Aardman Animations

                      Are they similar matrix