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?






algebra-precalculus polynomials factoring







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?






    algebra-precalculus polynomials factoring







    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?






      algebra-precalculus polynomials factoring







      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




      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
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          $begingroup$

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











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            2 Answers
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            2 Answers
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            active

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            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


















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