Find the value of $Q(x)$ at $x= -1$, knowing some of its properties.












2












$begingroup$



Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
$$
frac{Q(2x)}{Q(x+1)}= 8 -frac{56}{x+7},
quad forall x ne -7 text{ and } Q(x+1)neq 0,.
$$
Find $Q(-1)$.







algebra-precalculus polynomials rational-functions







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
    $$
    frac{Q(2x)}{Q(x+1)}= 8 -frac{56}{x+7},
    quad forall x ne -7 text{ and } Q(x+1)neq 0,.
    $$
    Find $Q(-1)$.







    algebra-precalculus polynomials rational-functions







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
      $$
      frac{Q(2x)}{Q(x+1)}= 8 -frac{56}{x+7},
      quad forall x ne -7 text{ and } Q(x+1)neq 0,.
      $$
      Find $Q(-1)$.







      algebra-precalculus polynomials rational-functions







      share|cite|improve this question











      $endgroup$





      Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
      $$
      frac{Q(2x)}{Q(x+1)}= 8 -frac{56}{x+7},
      quad forall x ne -7 text{ and } Q(x+1)neq 0,.
      $$
      Find $Q(-1)$.







      algebra-precalculus polynomials rational-functions




      algebra-precalculus polynomials rational-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 23:24









      Batominovski

      1




      1










      asked Dec 10 '18 at 22:43









      James CaldyJames Caldy

      191




      191






















          1 Answer
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          active

          oldest

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          4












          $begingroup$

          Rewrite the equation as
          $$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
          Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
          $$(x+3),R(2x)=2(x+1),R(x+1),.$$
          Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
          $$S(2x)=S(x+1),,$$
          which implies that $S(x)$ is a constant polynomial. Therefore,
          $$Q(x)=kx(x+2)(x+6)$$
          for some constant $k$.




          You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            Rewrite the equation as
            $$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
            Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
            $$(x+3),R(2x)=2(x+1),R(x+1),.$$
            Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
            $$S(2x)=S(x+1),,$$
            which implies that $S(x)$ is a constant polynomial. Therefore,
            $$Q(x)=kx(x+2)(x+6)$$
            for some constant $k$.




            You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              Rewrite the equation as
              $$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
              Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
              $$(x+3),R(2x)=2(x+1),R(x+1),.$$
              Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
              $$S(2x)=S(x+1),,$$
              which implies that $S(x)$ is a constant polynomial. Therefore,
              $$Q(x)=kx(x+2)(x+6)$$
              for some constant $k$.




              You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                Rewrite the equation as
                $$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
                Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
                $$(x+3),R(2x)=2(x+1),R(x+1),.$$
                Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
                $$S(2x)=S(x+1),,$$
                which implies that $S(x)$ is a constant polynomial. Therefore,
                $$Q(x)=kx(x+2)(x+6)$$
                for some constant $k$.




                You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.






                share|cite|improve this answer











                $endgroup$



                Rewrite the equation as
                $$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
                Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
                $$(x+3),R(2x)=2(x+1),R(x+1),.$$
                Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
                $$S(2x)=S(x+1),,$$
                which implies that $S(x)$ is a constant polynomial. Therefore,
                $$Q(x)=kx(x+2)(x+6)$$
                for some constant $k$.




                You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 10 '18 at 23:20

























                answered Dec 10 '18 at 23:13









                BatominovskiBatominovski

                1




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