How can we factor $ q^4 + q^2 + 1$












1












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Alright, I met this example solved in a book, but I want to know how did they come up with the answer:



$$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$



Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.






elementary-number-theory factoring







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  • $begingroup$
    here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
    $endgroup$
    – user25406
    Nov 8 '16 at 20:20
















1












$begingroup$


Alright, I met this example solved in a book, but I want to know how did they come up with the answer:



$$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$



Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.






elementary-number-theory factoring







share|cite|improve this question











$endgroup$












  • $begingroup$
    here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
    $endgroup$
    – user25406
    Nov 8 '16 at 20:20














1












1








1





$begingroup$


Alright, I met this example solved in a book, but I want to know how did they come up with the answer:



$$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$



Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.






elementary-number-theory factoring







share|cite|improve this question











$endgroup$




Alright, I met this example solved in a book, but I want to know how did they come up with the answer:



$$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$



Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.






elementary-number-theory factoring




elementary-number-theory factoring






share|cite|improve this question















share|cite|improve this question













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edited Jan 1 at 15:59









Bill Dubuque

211k29194648




211k29194648










asked Nov 8 '16 at 16:01









G. NickG. Nick

82




82












  • $begingroup$
    here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
    $endgroup$
    – user25406
    Nov 8 '16 at 20:20


















  • $begingroup$
    here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
    $endgroup$
    – user25406
    Nov 8 '16 at 20:20
















$begingroup$
here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
$endgroup$
– user25406
Nov 8 '16 at 20:20




$begingroup$
here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
$endgroup$
– user25406
Nov 8 '16 at 20:20










2 Answers
2






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oldest

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2












$begingroup$

Hint $ $ completing the square leads to a difference of squares, e.g.



$$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
end{eqnarray}$$ Here is another common example



$$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
&,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$






share|cite|improve this answer











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    1












    $begingroup$

    $$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
    Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
    $$(1+q^2)^2=1+q^4+2q^2$$
    Hence:
    $$1+q^2+q^4=(1+q^2)^2-q^2$$
    Which is the difference of two squares, something we know we can factor:
    $$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$




    Some often used rules for factoring are:
    $$(a+b)^2=a^2+b^2+2ab$$
    $$(a-b)(a+b)=a^2-b^2$$
    $$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






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      2 Answers
      2






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      2












      $begingroup$

      Hint $ $ completing the square leads to a difference of squares, e.g.



      $$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
      &,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
      end{eqnarray}$$       Here is another common example



      $$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
      &,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
      &,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Hint $ $ completing the square leads to a difference of squares, e.g.



        $$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
        &,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
        end{eqnarray}$$         Here is another common example



        $$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
        &,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
        &,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint $ $ completing the square leads to a difference of squares, e.g.



          $$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
          &,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
          end{eqnarray}$$           Here is another common example



          $$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
          &,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
          &,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$






          share|cite|improve this answer











          $endgroup$



          Hint $ $ completing the square leads to a difference of squares, e.g.



          $$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
          &,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
          end{eqnarray}$$           Here is another common example



          $$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
          &,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
          &,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 1 at 16:08

























          answered Nov 8 '16 at 16:05









          Bill DubuqueBill Dubuque

          211k29194648




          211k29194648























              1












              $begingroup$

              $$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
              Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
              $$(1+q^2)^2=1+q^4+2q^2$$
              Hence:
              $$1+q^2+q^4=(1+q^2)^2-q^2$$
              Which is the difference of two squares, something we know we can factor:
              $$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$




              Some often used rules for factoring are:
              $$(a+b)^2=a^2+b^2+2ab$$
              $$(a-b)(a+b)=a^2-b^2$$
              $$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
                Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
                $$(1+q^2)^2=1+q^4+2q^2$$
                Hence:
                $$1+q^2+q^4=(1+q^2)^2-q^2$$
                Which is the difference of two squares, something we know we can factor:
                $$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$




                Some often used rules for factoring are:
                $$(a+b)^2=a^2+b^2+2ab$$
                $$(a-b)(a+b)=a^2-b^2$$
                $$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
                  Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
                  $$(1+q^2)^2=1+q^4+2q^2$$
                  Hence:
                  $$1+q^2+q^4=(1+q^2)^2-q^2$$
                  Which is the difference of two squares, something we know we can factor:
                  $$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$




                  Some often used rules for factoring are:
                  $$(a+b)^2=a^2+b^2+2ab$$
                  $$(a-b)(a+b)=a^2-b^2$$
                  $$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$






                  share|cite|improve this answer









                  $endgroup$



                  $$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
                  Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
                  $$(1+q^2)^2=1+q^4+2q^2$$
                  Hence:
                  $$1+q^2+q^4=(1+q^2)^2-q^2$$
                  Which is the difference of two squares, something we know we can factor:
                  $$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$




                  Some often used rules for factoring are:
                  $$(a+b)^2=a^2+b^2+2ab$$
                  $$(a-b)(a+b)=a^2-b^2$$
                  $$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 8 '16 at 16:05









                  MastremMastrem

                  3,79711230




                  3,79711230






























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