Factor polynomials












1












$begingroup$


I am having trouble with these expressions:



$$x^4 - 23x^2 + 1$$
$$2y^2 - 5xy + 2x^2 - ay - ax - a^2$$
$$2x^3 - 4x^2y - x^2z + 2xy^2 - y^2z +2xyz$$



I tried to consult a chapter on factoring in my textbook. It seems to suggest the following:



Check for expressions like $a^n - b^n$ or $(a + b)^n$.



Remember that $x^2 + (a + b)x + ab =(x + a)(x + b)$.



Complete the square and check if any of the above apply.
(For example, $x^2 - 5x + 3$ add $pm (5/2)^2$ to get the expression equivalent to $(a^n + b^n)$.)



Grouping.



Regarding the first expression I tried the third method i.e. I added $(23/2)^2$ hoping to get the expression equivalent to $a^2 - b^2$.



In the second exercise I tried to factor monomial from different expressions to see some sort of pattern. I also tried to complete the square that is to add different expressions like $pm(5y/2)^2$. Formula itself reminds me of $(a + b + c)^2$. I am pretty sure the result is of the form $(a + b + c) (a - b + c)$ or something like that but I am still unsure how to proceed.



Regarding the third expression I don’t even know where to start. I tried grouping and other methods but I still can’t see a pattern.



Can someone give me a hint? Am I missing some rule? How to solve problems like that? What goes on in your head while you're looking at the expressions like that?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am having trouble with these expressions:



    $$x^4 - 23x^2 + 1$$
    $$2y^2 - 5xy + 2x^2 - ay - ax - a^2$$
    $$2x^3 - 4x^2y - x^2z + 2xy^2 - y^2z +2xyz$$



    I tried to consult a chapter on factoring in my textbook. It seems to suggest the following:



    Check for expressions like $a^n - b^n$ or $(a + b)^n$.



    Remember that $x^2 + (a + b)x + ab =(x + a)(x + b)$.



    Complete the square and check if any of the above apply.
    (For example, $x^2 - 5x + 3$ add $pm (5/2)^2$ to get the expression equivalent to $(a^n + b^n)$.)



    Grouping.



    Regarding the first expression I tried the third method i.e. I added $(23/2)^2$ hoping to get the expression equivalent to $a^2 - b^2$.



    In the second exercise I tried to factor monomial from different expressions to see some sort of pattern. I also tried to complete the square that is to add different expressions like $pm(5y/2)^2$. Formula itself reminds me of $(a + b + c)^2$. I am pretty sure the result is of the form $(a + b + c) (a - b + c)$ or something like that but I am still unsure how to proceed.



    Regarding the third expression I don’t even know where to start. I tried grouping and other methods but I still can’t see a pattern.



    Can someone give me a hint? Am I missing some rule? How to solve problems like that? What goes on in your head while you're looking at the expressions like that?










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I am having trouble with these expressions:



      $$x^4 - 23x^2 + 1$$
      $$2y^2 - 5xy + 2x^2 - ay - ax - a^2$$
      $$2x^3 - 4x^2y - x^2z + 2xy^2 - y^2z +2xyz$$



      I tried to consult a chapter on factoring in my textbook. It seems to suggest the following:



      Check for expressions like $a^n - b^n$ or $(a + b)^n$.



      Remember that $x^2 + (a + b)x + ab =(x + a)(x + b)$.



      Complete the square and check if any of the above apply.
      (For example, $x^2 - 5x + 3$ add $pm (5/2)^2$ to get the expression equivalent to $(a^n + b^n)$.)



      Grouping.



      Regarding the first expression I tried the third method i.e. I added $(23/2)^2$ hoping to get the expression equivalent to $a^2 - b^2$.



      In the second exercise I tried to factor monomial from different expressions to see some sort of pattern. I also tried to complete the square that is to add different expressions like $pm(5y/2)^2$. Formula itself reminds me of $(a + b + c)^2$. I am pretty sure the result is of the form $(a + b + c) (a - b + c)$ or something like that but I am still unsure how to proceed.



      Regarding the third expression I don’t even know where to start. I tried grouping and other methods but I still can’t see a pattern.



      Can someone give me a hint? Am I missing some rule? How to solve problems like that? What goes on in your head while you're looking at the expressions like that?










      share|cite|improve this question











      $endgroup$




      I am having trouble with these expressions:



      $$x^4 - 23x^2 + 1$$
      $$2y^2 - 5xy + 2x^2 - ay - ax - a^2$$
      $$2x^3 - 4x^2y - x^2z + 2xy^2 - y^2z +2xyz$$



      I tried to consult a chapter on factoring in my textbook. It seems to suggest the following:



      Check for expressions like $a^n - b^n$ or $(a + b)^n$.



      Remember that $x^2 + (a + b)x + ab =(x + a)(x + b)$.



      Complete the square and check if any of the above apply.
      (For example, $x^2 - 5x + 3$ add $pm (5/2)^2$ to get the expression equivalent to $(a^n + b^n)$.)



      Grouping.



      Regarding the first expression I tried the third method i.e. I added $(23/2)^2$ hoping to get the expression equivalent to $a^2 - b^2$.



      In the second exercise I tried to factor monomial from different expressions to see some sort of pattern. I also tried to complete the square that is to add different expressions like $pm(5y/2)^2$. Formula itself reminds me of $(a + b + c)^2$. I am pretty sure the result is of the form $(a + b + c) (a - b + c)$ or something like that but I am still unsure how to proceed.



      Regarding the third expression I don’t even know where to start. I tried grouping and other methods but I still can’t see a pattern.



      Can someone give me a hint? Am I missing some rule? How to solve problems like that? What goes on in your head while you're looking at the expressions like that?







      algebra-precalculus polynomials factoring






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 31 '18 at 18:31









      David G. Stork

      11k41432




      11k41432










      asked Dec 31 '18 at 17:14









      I.am.bad.at.factoringI.am.bad.at.factoring

      84




      84






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          $$x^4-23x^2+1=x^4+2x^2+1-25x^2=(x^2-5x+1)(x^2+5x+1).$$
          $$2x^2-5xy+2y^2-ax-ay-a^2=$$
          $$=2x^2-5xy+2y^2+frac{1}{4}(x+y)^2-frac{1}{4}(x+y)^2-a(x+y)-a^2=$$
          $$=frac{9}{4}(x-y)^2-left(frac{1}{2}(x+y)+aright)^2=$$
          $$=left(frac{3}{2}(x-y)-frac{1}{2}(x+y)-aright)left(frac{3}{2}(x-y)+frac{1}{2}(x+y)+aright)=$$
          $$=(x-2y-a)(2x-y+a).$$
          The third is irreducible.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Substitute $$x^2=t$$ in the first one and solve the quadratic, then you can factorize.Hint for the second: $$y-2x-a$$ is one factor.
            Hint for the last one: One factor is $$(x-y)^2$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I am curious how you determined the factor for the last one. Thanks.
              $endgroup$
              – paw88789
              Dec 31 '18 at 18:20











            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%2f3057871%2ffactor-polynomials%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












            $begingroup$

            $$x^4-23x^2+1=x^4+2x^2+1-25x^2=(x^2-5x+1)(x^2+5x+1).$$
            $$2x^2-5xy+2y^2-ax-ay-a^2=$$
            $$=2x^2-5xy+2y^2+frac{1}{4}(x+y)^2-frac{1}{4}(x+y)^2-a(x+y)-a^2=$$
            $$=frac{9}{4}(x-y)^2-left(frac{1}{2}(x+y)+aright)^2=$$
            $$=left(frac{3}{2}(x-y)-frac{1}{2}(x+y)-aright)left(frac{3}{2}(x-y)+frac{1}{2}(x+y)+aright)=$$
            $$=(x-2y-a)(2x-y+a).$$
            The third is irreducible.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              $$x^4-23x^2+1=x^4+2x^2+1-25x^2=(x^2-5x+1)(x^2+5x+1).$$
              $$2x^2-5xy+2y^2-ax-ay-a^2=$$
              $$=2x^2-5xy+2y^2+frac{1}{4}(x+y)^2-frac{1}{4}(x+y)^2-a(x+y)-a^2=$$
              $$=frac{9}{4}(x-y)^2-left(frac{1}{2}(x+y)+aright)^2=$$
              $$=left(frac{3}{2}(x-y)-frac{1}{2}(x+y)-aright)left(frac{3}{2}(x-y)+frac{1}{2}(x+y)+aright)=$$
              $$=(x-2y-a)(2x-y+a).$$
              The third is irreducible.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                $$x^4-23x^2+1=x^4+2x^2+1-25x^2=(x^2-5x+1)(x^2+5x+1).$$
                $$2x^2-5xy+2y^2-ax-ay-a^2=$$
                $$=2x^2-5xy+2y^2+frac{1}{4}(x+y)^2-frac{1}{4}(x+y)^2-a(x+y)-a^2=$$
                $$=frac{9}{4}(x-y)^2-left(frac{1}{2}(x+y)+aright)^2=$$
                $$=left(frac{3}{2}(x-y)-frac{1}{2}(x+y)-aright)left(frac{3}{2}(x-y)+frac{1}{2}(x+y)+aright)=$$
                $$=(x-2y-a)(2x-y+a).$$
                The third is irreducible.






                share|cite|improve this answer











                $endgroup$



                $$x^4-23x^2+1=x^4+2x^2+1-25x^2=(x^2-5x+1)(x^2+5x+1).$$
                $$2x^2-5xy+2y^2-ax-ay-a^2=$$
                $$=2x^2-5xy+2y^2+frac{1}{4}(x+y)^2-frac{1}{4}(x+y)^2-a(x+y)-a^2=$$
                $$=frac{9}{4}(x-y)^2-left(frac{1}{2}(x+y)+aright)^2=$$
                $$=left(frac{3}{2}(x-y)-frac{1}{2}(x+y)-aright)left(frac{3}{2}(x-y)+frac{1}{2}(x+y)+aright)=$$
                $$=(x-2y-a)(2x-y+a).$$
                The third is irreducible.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 31 '18 at 17:47

























                answered Dec 31 '18 at 17:33









                Michael RozenbergMichael Rozenberg

                105k1893198




                105k1893198























                    0












                    $begingroup$

                    Substitute $$x^2=t$$ in the first one and solve the quadratic, then you can factorize.Hint for the second: $$y-2x-a$$ is one factor.
                    Hint for the last one: One factor is $$(x-y)^2$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      I am curious how you determined the factor for the last one. Thanks.
                      $endgroup$
                      – paw88789
                      Dec 31 '18 at 18:20
















                    0












                    $begingroup$

                    Substitute $$x^2=t$$ in the first one and solve the quadratic, then you can factorize.Hint for the second: $$y-2x-a$$ is one factor.
                    Hint for the last one: One factor is $$(x-y)^2$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      I am curious how you determined the factor for the last one. Thanks.
                      $endgroup$
                      – paw88789
                      Dec 31 '18 at 18:20














                    0












                    0








                    0





                    $begingroup$

                    Substitute $$x^2=t$$ in the first one and solve the quadratic, then you can factorize.Hint for the second: $$y-2x-a$$ is one factor.
                    Hint for the last one: One factor is $$(x-y)^2$$






                    share|cite|improve this answer











                    $endgroup$



                    Substitute $$x^2=t$$ in the first one and solve the quadratic, then you can factorize.Hint for the second: $$y-2x-a$$ is one factor.
                    Hint for the last one: One factor is $$(x-y)^2$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 31 '18 at 17:27

























                    answered Dec 31 '18 at 17:22









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    76.5k42866




                    76.5k42866












                    • $begingroup$
                      I am curious how you determined the factor for the last one. Thanks.
                      $endgroup$
                      – paw88789
                      Dec 31 '18 at 18:20


















                    • $begingroup$
                      I am curious how you determined the factor for the last one. Thanks.
                      $endgroup$
                      – paw88789
                      Dec 31 '18 at 18:20
















                    $begingroup$
                    I am curious how you determined the factor for the last one. Thanks.
                    $endgroup$
                    – paw88789
                    Dec 31 '18 at 18:20




                    $begingroup$
                    I am curious how you determined the factor for the last one. Thanks.
                    $endgroup$
                    – paw88789
                    Dec 31 '18 at 18:20


















                    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%2f3057871%2ffactor-polynomials%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

                    Ellipse (mathématiques)

                    Quarter-circle Tiles

                    Mont Emei