Binomial expansion involving partial fractions












0












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Sorry I do not know how to use the formatting will try my best.



Q. Find the binomial expansion up to $x^2$ of:



$$frac{3+2x^2}{(2x+1)(x-3)^2}$$



For the partial fraction I get:



$$frac{2}{7}frac{1}{2x+1} + frac{6}{7}frac{1}{x-3} + frac{3}{(x-3)^2}$$



Then I did the following:



$$(2x+1)^{-1} = 1 - 2x + 4x^2$$



$$(x-3)^{-1}= frac{1}{3} + frac{x}{9} + frac{x^2}{27}$$



$$(x-3)^{-2} = frac{1}{3} + frac{2x}{9} + frac{x^2}{9}$$



When I add them I get completely the wrong answer:



Correct answer is
$$frac{1}{3} + frac{4x}{9} + frac{11x^2}{9}.$$






partial-fractions







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  • $begingroup$
    Do you polynomial division by increasing powers?
    $endgroup$
    – Bernard
    Nov 21 '17 at 15:32
















0












$begingroup$


Sorry I do not know how to use the formatting will try my best.



Q. Find the binomial expansion up to $x^2$ of:



$$frac{3+2x^2}{(2x+1)(x-3)^2}$$



For the partial fraction I get:



$$frac{2}{7}frac{1}{2x+1} + frac{6}{7}frac{1}{x-3} + frac{3}{(x-3)^2}$$



Then I did the following:



$$(2x+1)^{-1} = 1 - 2x + 4x^2$$



$$(x-3)^{-1}= frac{1}{3} + frac{x}{9} + frac{x^2}{27}$$



$$(x-3)^{-2} = frac{1}{3} + frac{2x}{9} + frac{x^2}{9}$$



When I add them I get completely the wrong answer:



Correct answer is
$$frac{1}{3} + frac{4x}{9} + frac{11x^2}{9}.$$






partial-fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you polynomial division by increasing powers?
    $endgroup$
    – Bernard
    Nov 21 '17 at 15:32














0












0








0





$begingroup$


Sorry I do not know how to use the formatting will try my best.



Q. Find the binomial expansion up to $x^2$ of:



$$frac{3+2x^2}{(2x+1)(x-3)^2}$$



For the partial fraction I get:



$$frac{2}{7}frac{1}{2x+1} + frac{6}{7}frac{1}{x-3} + frac{3}{(x-3)^2}$$



Then I did the following:



$$(2x+1)^{-1} = 1 - 2x + 4x^2$$



$$(x-3)^{-1}= frac{1}{3} + frac{x}{9} + frac{x^2}{27}$$



$$(x-3)^{-2} = frac{1}{3} + frac{2x}{9} + frac{x^2}{9}$$



When I add them I get completely the wrong answer:



Correct answer is
$$frac{1}{3} + frac{4x}{9} + frac{11x^2}{9}.$$






partial-fractions







share|cite|improve this question











$endgroup$




Sorry I do not know how to use the formatting will try my best.



Q. Find the binomial expansion up to $x^2$ of:



$$frac{3+2x^2}{(2x+1)(x-3)^2}$$



For the partial fraction I get:



$$frac{2}{7}frac{1}{2x+1} + frac{6}{7}frac{1}{x-3} + frac{3}{(x-3)^2}$$



Then I did the following:



$$(2x+1)^{-1} = 1 - 2x + 4x^2$$



$$(x-3)^{-1}= frac{1}{3} + frac{x}{9} + frac{x^2}{27}$$



$$(x-3)^{-2} = frac{1}{3} + frac{2x}{9} + frac{x^2}{9}$$



When I add them I get completely the wrong answer:



Correct answer is
$$frac{1}{3} + frac{4x}{9} + frac{11x^2}{9}.$$






partial-fractions




partial-fractions






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share|cite|improve this question













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share|cite|improve this question








edited Nov 21 '17 at 15:44







boi Shift

















asked Nov 21 '17 at 15:19









boi Shiftboi Shift

266




266












  • $begingroup$
    Do you polynomial division by increasing powers?
    $endgroup$
    – Bernard
    Nov 21 '17 at 15:32


















  • $begingroup$
    Do you polynomial division by increasing powers?
    $endgroup$
    – Bernard
    Nov 21 '17 at 15:32
















$begingroup$
Do you polynomial division by increasing powers?
$endgroup$
– Bernard
Nov 21 '17 at 15:32




$begingroup$
Do you polynomial division by increasing powers?
$endgroup$
– Bernard
Nov 21 '17 at 15:32










3 Answers
3






active

oldest

votes


















1












$begingroup$

You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
$$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):



begin{alignat}{6}
& & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
& & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
& & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
&&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
&&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
&&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
&&&&-dfrac{182}{9}x^3&+dotsm
end{alignat}






share|cite|improve this answer









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    0












    $begingroup$

    Hint:



    How about finding $a,b,c$ in



    $$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$



    by comparing the coefficients of different powers of $x$



    For example, $(-3)^2a=3$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
      $endgroup$
      – boi Shift
      Nov 21 '17 at 15:45



















    0












    $begingroup$

    $$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
      $$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
      We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):



      begin{alignat}{6}
      & & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
      9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
      & & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
      & & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
      &&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
      &&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
      &&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
      &&&&-dfrac{182}{9}x^3&+dotsm
      end{alignat}






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
        $$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
        We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):



        begin{alignat}{6}
        & & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
        9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
        & & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
        & & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
        &&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
        &&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
        &&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
        &&&&-dfrac{182}{9}x^3&+dotsm
        end{alignat}






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
          $$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
          We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):



          begin{alignat}{6}
          & & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
          9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
          & & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
          & & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
          &&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
          &&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
          &&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
          &&&&-dfrac{182}{9}x^3&+dotsm
          end{alignat}






          share|cite|improve this answer









          $endgroup$



          You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
          $$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
          We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):



          begin{alignat}{6}
          & & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
          9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
          & & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
          & & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
          &&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
          &&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
          &&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
          &&&&-dfrac{182}{9}x^3&+dotsm
          end{alignat}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 21 '17 at 17:15









          BernardBernard

          120k740113




          120k740113























              0












              $begingroup$

              Hint:



              How about finding $a,b,c$ in



              $$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$



              by comparing the coefficients of different powers of $x$



              For example, $(-3)^2a=3$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
                $endgroup$
                – boi Shift
                Nov 21 '17 at 15:45
















              0












              $begingroup$

              Hint:



              How about finding $a,b,c$ in



              $$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$



              by comparing the coefficients of different powers of $x$



              For example, $(-3)^2a=3$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
                $endgroup$
                – boi Shift
                Nov 21 '17 at 15:45














              0












              0








              0





              $begingroup$

              Hint:



              How about finding $a,b,c$ in



              $$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$



              by comparing the coefficients of different powers of $x$



              For example, $(-3)^2a=3$






              share|cite|improve this answer









              $endgroup$



              Hint:



              How about finding $a,b,c$ in



              $$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$



              by comparing the coefficients of different powers of $x$



              For example, $(-3)^2a=3$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 21 '17 at 15:33









              lab bhattacharjeelab bhattacharjee

              225k15157275




              225k15157275












              • $begingroup$
                The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
                $endgroup$
                – boi Shift
                Nov 21 '17 at 15:45


















              • $begingroup$
                The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
                $endgroup$
                – boi Shift
                Nov 21 '17 at 15:45
















              $begingroup$
              The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
              $endgroup$
              – boi Shift
              Nov 21 '17 at 15:45




              $begingroup$
              The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
              $endgroup$
              – boi Shift
              Nov 21 '17 at 15:45











              0












              $begingroup$

              $$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$






                  share|cite|improve this answer









                  $endgroup$



                  $$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 '17 at 15:50









                  lab bhattacharjeelab bhattacharjee

                  225k15157275




                  225k15157275






























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