Find p and q of this polynomial division [duplicate]












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  • $ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$

    9 answers




Dividing $x^4 + px^3 + qx^2 - 16x -12 \$ by $(x+1)(x+3)$, the remainder is $2x+3$, find $p$ and $q$



I dont know how to solve this, please i need help



regards.






algebra-precalculus polynomials







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marked as duplicate by Bill Dubuque algebra-precalculus
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 Jan 5 at 21:12


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    $begingroup$



    This question already has an answer here:




    • $ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$

      9 answers




    Dividing $x^4 + px^3 + qx^2 - 16x -12 \$ by $(x+1)(x+3)$, the remainder is $2x+3$, find $p$ and $q$



    I dont know how to solve this, please i need help



    regards.






    algebra-precalculus polynomials







    share|cite|improve this question











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    marked as duplicate by Bill Dubuque algebra-precalculus
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     Jan 5 at 21:12


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      0





      $begingroup$



      This question already has an answer here:




      • $ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$

        9 answers




      Dividing $x^4 + px^3 + qx^2 - 16x -12 \$ by $(x+1)(x+3)$, the remainder is $2x+3$, find $p$ and $q$



      I dont know how to solve this, please i need help



      regards.






      algebra-precalculus polynomials







      share|cite|improve this question











      $endgroup$





      This question already has an answer here:




      • $ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$

        9 answers




      Dividing $x^4 + px^3 + qx^2 - 16x -12 \$ by $(x+1)(x+3)$, the remainder is $2x+3$, find $p$ and $q$



      I dont know how to solve this, please i need help



      regards.





      This question already has an answer here:




      • $ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$

        9 answers







      algebra-precalculus polynomials




      algebra-precalculus polynomials






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      share|cite|improve this question













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      edited Jan 5 at 19:27









      José Carlos Santos

      153k22123226




      153k22123226










      asked Jan 5 at 19:20









      englishworkvgsenglishworkvgs

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      marked as duplicate by Bill Dubuque algebra-precalculus
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       Jan 5 at 21:12


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          2 Answers
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          $begingroup$

          First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$



          begin{align}
          q(x) &= p(x) - (2x+3) \
          &= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
          &= x^4 + px^3 + qx^2 -18x -15
          end{align}



          Now, $(x+1)|q(x)$ and $(x+3)|q(x)$



          Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$






          share|cite|improve this answer









          $endgroup$





















            7












            $begingroup$

            You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?






            share|cite|improve this answer











            $endgroup$




















              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$



              begin{align}
              q(x) &= p(x) - (2x+3) \
              &= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
              &= x^4 + px^3 + qx^2 -18x -15
              end{align}



              Now, $(x+1)|q(x)$ and $(x+3)|q(x)$



              Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$



                begin{align}
                q(x) &= p(x) - (2x+3) \
                &= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
                &= x^4 + px^3 + qx^2 -18x -15
                end{align}



                Now, $(x+1)|q(x)$ and $(x+3)|q(x)$



                Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$



                  begin{align}
                  q(x) &= p(x) - (2x+3) \
                  &= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
                  &= x^4 + px^3 + qx^2 -18x -15
                  end{align}



                  Now, $(x+1)|q(x)$ and $(x+3)|q(x)$



                  Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$






                  share|cite|improve this answer









                  $endgroup$



                  First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$



                  begin{align}
                  q(x) &= p(x) - (2x+3) \
                  &= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
                  &= x^4 + px^3 + qx^2 -18x -15
                  end{align}



                  Now, $(x+1)|q(x)$ and $(x+3)|q(x)$



                  Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 20:19









                  kelalakakelalaka

                  329212




                  329212























                      7












                      $begingroup$

                      You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?






                      share|cite|improve this answer











                      $endgroup$


















                        7












                        $begingroup$

                        You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?






                        share|cite|improve this answer











                        $endgroup$
















                          7












                          7








                          7





                          $begingroup$

                          You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?






                          share|cite|improve this answer











                          $endgroup$



                          You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 6 at 8:01

























                          answered Jan 5 at 19:25









                          José Carlos SantosJosé Carlos Santos

                          153k22123226




                          153k22123226















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