Polynomial division with remainder










2















If the polynomial
$$x^4-x^3+ax^2+bx+c$$
divided by the polynomial
$$x^3+2x^2-3x+1$$
gives the remainder $$3x^2-2x+1$$
then how much is (a+b)c?




So what I know, and how I solved these problems before, I can write this down like this:



$x^4-x^3+ax^2+bx+c=Q(x)(x^3+2x^2-3x+1)+3x^2-2x+1$



And from here I should get the roots of $x^3+2x^2-3x+1$ so I could use the remainder only since I don't know $Q(x)$



$x^3+2x^2-3x+1=0$



but I can't find the roots of this by hand, so I used WolframAlpha to get the roots, but they are not whole numbers so I couldn't write this in the form of $(x-x_1)(x-x_2)(x-x_3)$ and then just use $P(x_1)$, $P(x_2)$, $P(x_3)$ to get $a,b,$ and $c$



I tried regular dividing on these polynomials to get $Q(x)$ but I couldn't divide them because of the $a,b$ and $c$ present










share|cite|improve this question


























    2















    If the polynomial
    $$x^4-x^3+ax^2+bx+c$$
    divided by the polynomial
    $$x^3+2x^2-3x+1$$
    gives the remainder $$3x^2-2x+1$$
    then how much is (a+b)c?




    So what I know, and how I solved these problems before, I can write this down like this:



    $x^4-x^3+ax^2+bx+c=Q(x)(x^3+2x^2-3x+1)+3x^2-2x+1$



    And from here I should get the roots of $x^3+2x^2-3x+1$ so I could use the remainder only since I don't know $Q(x)$



    $x^3+2x^2-3x+1=0$



    but I can't find the roots of this by hand, so I used WolframAlpha to get the roots, but they are not whole numbers so I couldn't write this in the form of $(x-x_1)(x-x_2)(x-x_3)$ and then just use $P(x_1)$, $P(x_2)$, $P(x_3)$ to get $a,b,$ and $c$



    I tried regular dividing on these polynomials to get $Q(x)$ but I couldn't divide them because of the $a,b$ and $c$ present










    share|cite|improve this question
























      2












      2








      2








      If the polynomial
      $$x^4-x^3+ax^2+bx+c$$
      divided by the polynomial
      $$x^3+2x^2-3x+1$$
      gives the remainder $$3x^2-2x+1$$
      then how much is (a+b)c?




      So what I know, and how I solved these problems before, I can write this down like this:



      $x^4-x^3+ax^2+bx+c=Q(x)(x^3+2x^2-3x+1)+3x^2-2x+1$



      And from here I should get the roots of $x^3+2x^2-3x+1$ so I could use the remainder only since I don't know $Q(x)$



      $x^3+2x^2-3x+1=0$



      but I can't find the roots of this by hand, so I used WolframAlpha to get the roots, but they are not whole numbers so I couldn't write this in the form of $(x-x_1)(x-x_2)(x-x_3)$ and then just use $P(x_1)$, $P(x_2)$, $P(x_3)$ to get $a,b,$ and $c$



      I tried regular dividing on these polynomials to get $Q(x)$ but I couldn't divide them because of the $a,b$ and $c$ present










      share|cite|improve this question














      If the polynomial
      $$x^4-x^3+ax^2+bx+c$$
      divided by the polynomial
      $$x^3+2x^2-3x+1$$
      gives the remainder $$3x^2-2x+1$$
      then how much is (a+b)c?




      So what I know, and how I solved these problems before, I can write this down like this:



      $x^4-x^3+ax^2+bx+c=Q(x)(x^3+2x^2-3x+1)+3x^2-2x+1$



      And from here I should get the roots of $x^3+2x^2-3x+1$ so I could use the remainder only since I don't know $Q(x)$



      $x^3+2x^2-3x+1=0$



      but I can't find the roots of this by hand, so I used WolframAlpha to get the roots, but they are not whole numbers so I couldn't write this in the form of $(x-x_1)(x-x_2)(x-x_3)$ and then just use $P(x_1)$, $P(x_2)$, $P(x_3)$ to get $a,b,$ and $c$



      I tried regular dividing on these polynomials to get $Q(x)$ but I couldn't divide them because of the $a,b$ and $c$ present







      polynomials






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 11 at 10:40









      Aleksa

      33612




      33612




















          3 Answers
          3






          active

          oldest

          votes


















          2














          Let (original) polynomial $P(x) = x^4 - x^3 + ax^2 + bx + c$ and divisor polynomial $D(x) = x^3 + 2x^2 - 3x + 1$. The remainder polynomial $R(x) = 3x^2 - 2x + 1$. We can also define quotient polynomial $Q(x) = alpha x + beta$.



          We have that $P(x) = Q(x)cdot D(x) + R(x)$.



          By inspection of the coefficient of $x^4$ on both sides, $alpha = 1$, i.e. $Q(x)$ is monic.



          Considering the coefficient of $x^3$ on both sides, we have that $-x^3 = 2x^3 + beta x^3$, giving $beta = -3$. So $Q(x) = (x-3)$.



          By now substituting convenient values for $x$, we can solve for $a, b, c$ quite easily. But we don't have to go all the way on that, since we only need to find $(a+b)c$.



          We can find $c$ by putting $x = 0$, giving $c = (-3)(1) + 1 = -2$.



          We can find $a+b$ by putting $x = 1$, giving $1-1 + a + b - 2 = (-2)(1+2-3+1) + 3-2+1$, giving $a+b = 2$.



          So $(a+b)c = 2(-2) = -4$.






          share|cite|improve this answer






















          • Man math is beautiful! Could you explain just the part of $x^3$ where you get $beta$ in more detail?
            – Aleksa
            Nov 11 at 11:35







          • 1




            @Aleksa I agree! :) What I wanted to add is that, for remainder and factor theorem (and polynomial division in general), I often find it useful to go back to the basic theory because it's quite obvious and often very easy when you write it out. Better not to learn too many things by rote.
            – Deepak
            Nov 11 at 11:37






          • 1




            Sorry, missed your edit. There are only two "ways" of getting the $x^3$ term of $P(x)$ on the LHS from the expression on the RHS - multiplying the $x^3$ term of $D(x)$ by the constant term of $Q(x)$ and multiplying the $x^2$ term of $D(x)$ by the $x$ term of $Q(x)$. Adding these (signed) terms up on both sides gives you a quick linear equation you can solve for $beta$.
            – Deepak
            Nov 11 at 11:39











          • Thanks for your feedback!
            – Aleksa
            Nov 11 at 11:46










          • Sure, glad to help! :)
            – Deepak
            Nov 11 at 11:46


















          3














          Hint: Compute the remainder of $x^4-x^3$ when divided by $x^3+2x^2-3x+1$.






          share|cite|improve this answer




















          • I've got $ax^2+bx+c=(x^3+2x^2-3x+1)[Q(x)-(x+3)]-6x^2+8x+4$ now, what do I do from this point?
            – Aleksa
            Nov 11 at 10:59










          • The remainder was $9x^2-10x-3$, so I moved $(x^3-2x^2-3x+1)(x-3)+9x^2-10x-3$ on the other side to get that
            – Aleksa
            Nov 11 at 11:00



















          0














          Maybe if you go through the steps for polynomial long division you will get an answer for the remainder in terms of a and b. Then you can equate the x terms with the same order, to determine a and b.






          share|cite|improve this answer




















            Your Answer





            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader:
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            ,
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













            draft saved

            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2993694%2fpolynomial-division-with-remainder%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            Let (original) polynomial $P(x) = x^4 - x^3 + ax^2 + bx + c$ and divisor polynomial $D(x) = x^3 + 2x^2 - 3x + 1$. The remainder polynomial $R(x) = 3x^2 - 2x + 1$. We can also define quotient polynomial $Q(x) = alpha x + beta$.



            We have that $P(x) = Q(x)cdot D(x) + R(x)$.



            By inspection of the coefficient of $x^4$ on both sides, $alpha = 1$, i.e. $Q(x)$ is monic.



            Considering the coefficient of $x^3$ on both sides, we have that $-x^3 = 2x^3 + beta x^3$, giving $beta = -3$. So $Q(x) = (x-3)$.



            By now substituting convenient values for $x$, we can solve for $a, b, c$ quite easily. But we don't have to go all the way on that, since we only need to find $(a+b)c$.



            We can find $c$ by putting $x = 0$, giving $c = (-3)(1) + 1 = -2$.



            We can find $a+b$ by putting $x = 1$, giving $1-1 + a + b - 2 = (-2)(1+2-3+1) + 3-2+1$, giving $a+b = 2$.



            So $(a+b)c = 2(-2) = -4$.






            share|cite|improve this answer






















            • Man math is beautiful! Could you explain just the part of $x^3$ where you get $beta$ in more detail?
              – Aleksa
              Nov 11 at 11:35







            • 1




              @Aleksa I agree! :) What I wanted to add is that, for remainder and factor theorem (and polynomial division in general), I often find it useful to go back to the basic theory because it's quite obvious and often very easy when you write it out. Better not to learn too many things by rote.
              – Deepak
              Nov 11 at 11:37






            • 1




              Sorry, missed your edit. There are only two "ways" of getting the $x^3$ term of $P(x)$ on the LHS from the expression on the RHS - multiplying the $x^3$ term of $D(x)$ by the constant term of $Q(x)$ and multiplying the $x^2$ term of $D(x)$ by the $x$ term of $Q(x)$. Adding these (signed) terms up on both sides gives you a quick linear equation you can solve for $beta$.
              – Deepak
              Nov 11 at 11:39











            • Thanks for your feedback!
              – Aleksa
              Nov 11 at 11:46










            • Sure, glad to help! :)
              – Deepak
              Nov 11 at 11:46















            2














            Let (original) polynomial $P(x) = x^4 - x^3 + ax^2 + bx + c$ and divisor polynomial $D(x) = x^3 + 2x^2 - 3x + 1$. The remainder polynomial $R(x) = 3x^2 - 2x + 1$. We can also define quotient polynomial $Q(x) = alpha x + beta$.



            We have that $P(x) = Q(x)cdot D(x) + R(x)$.



            By inspection of the coefficient of $x^4$ on both sides, $alpha = 1$, i.e. $Q(x)$ is monic.



            Considering the coefficient of $x^3$ on both sides, we have that $-x^3 = 2x^3 + beta x^3$, giving $beta = -3$. So $Q(x) = (x-3)$.



            By now substituting convenient values for $x$, we can solve for $a, b, c$ quite easily. But we don't have to go all the way on that, since we only need to find $(a+b)c$.



            We can find $c$ by putting $x = 0$, giving $c = (-3)(1) + 1 = -2$.



            We can find $a+b$ by putting $x = 1$, giving $1-1 + a + b - 2 = (-2)(1+2-3+1) + 3-2+1$, giving $a+b = 2$.



            So $(a+b)c = 2(-2) = -4$.






            share|cite|improve this answer






















            • Man math is beautiful! Could you explain just the part of $x^3$ where you get $beta$ in more detail?
              – Aleksa
              Nov 11 at 11:35







            • 1




              @Aleksa I agree! :) What I wanted to add is that, for remainder and factor theorem (and polynomial division in general), I often find it useful to go back to the basic theory because it's quite obvious and often very easy when you write it out. Better not to learn too many things by rote.
              – Deepak
              Nov 11 at 11:37






            • 1




              Sorry, missed your edit. There are only two "ways" of getting the $x^3$ term of $P(x)$ on the LHS from the expression on the RHS - multiplying the $x^3$ term of $D(x)$ by the constant term of $Q(x)$ and multiplying the $x^2$ term of $D(x)$ by the $x$ term of $Q(x)$. Adding these (signed) terms up on both sides gives you a quick linear equation you can solve for $beta$.
              – Deepak
              Nov 11 at 11:39











            • Thanks for your feedback!
              – Aleksa
              Nov 11 at 11:46










            • Sure, glad to help! :)
              – Deepak
              Nov 11 at 11:46













            2












            2








            2






            Let (original) polynomial $P(x) = x^4 - x^3 + ax^2 + bx + c$ and divisor polynomial $D(x) = x^3 + 2x^2 - 3x + 1$. The remainder polynomial $R(x) = 3x^2 - 2x + 1$. We can also define quotient polynomial $Q(x) = alpha x + beta$.



            We have that $P(x) = Q(x)cdot D(x) + R(x)$.



            By inspection of the coefficient of $x^4$ on both sides, $alpha = 1$, i.e. $Q(x)$ is monic.



            Considering the coefficient of $x^3$ on both sides, we have that $-x^3 = 2x^3 + beta x^3$, giving $beta = -3$. So $Q(x) = (x-3)$.



            By now substituting convenient values for $x$, we can solve for $a, b, c$ quite easily. But we don't have to go all the way on that, since we only need to find $(a+b)c$.



            We can find $c$ by putting $x = 0$, giving $c = (-3)(1) + 1 = -2$.



            We can find $a+b$ by putting $x = 1$, giving $1-1 + a + b - 2 = (-2)(1+2-3+1) + 3-2+1$, giving $a+b = 2$.



            So $(a+b)c = 2(-2) = -4$.






            share|cite|improve this answer














            Let (original) polynomial $P(x) = x^4 - x^3 + ax^2 + bx + c$ and divisor polynomial $D(x) = x^3 + 2x^2 - 3x + 1$. The remainder polynomial $R(x) = 3x^2 - 2x + 1$. We can also define quotient polynomial $Q(x) = alpha x + beta$.



            We have that $P(x) = Q(x)cdot D(x) + R(x)$.



            By inspection of the coefficient of $x^4$ on both sides, $alpha = 1$, i.e. $Q(x)$ is monic.



            Considering the coefficient of $x^3$ on both sides, we have that $-x^3 = 2x^3 + beta x^3$, giving $beta = -3$. So $Q(x) = (x-3)$.



            By now substituting convenient values for $x$, we can solve for $a, b, c$ quite easily. But we don't have to go all the way on that, since we only need to find $(a+b)c$.



            We can find $c$ by putting $x = 0$, giving $c = (-3)(1) + 1 = -2$.



            We can find $a+b$ by putting $x = 1$, giving $1-1 + a + b - 2 = (-2)(1+2-3+1) + 3-2+1$, giving $a+b = 2$.



            So $(a+b)c = 2(-2) = -4$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 11 at 11:34

























            answered Nov 11 at 11:27









            Deepak

            16.6k11436




            16.6k11436











            • Man math is beautiful! Could you explain just the part of $x^3$ where you get $beta$ in more detail?
              – Aleksa
              Nov 11 at 11:35







            • 1




              @Aleksa I agree! :) What I wanted to add is that, for remainder and factor theorem (and polynomial division in general), I often find it useful to go back to the basic theory because it's quite obvious and often very easy when you write it out. Better not to learn too many things by rote.
              – Deepak
              Nov 11 at 11:37






            • 1




              Sorry, missed your edit. There are only two "ways" of getting the $x^3$ term of $P(x)$ on the LHS from the expression on the RHS - multiplying the $x^3$ term of $D(x)$ by the constant term of $Q(x)$ and multiplying the $x^2$ term of $D(x)$ by the $x$ term of $Q(x)$. Adding these (signed) terms up on both sides gives you a quick linear equation you can solve for $beta$.
              – Deepak
              Nov 11 at 11:39











            • Thanks for your feedback!
              – Aleksa
              Nov 11 at 11:46










            • Sure, glad to help! :)
              – Deepak
              Nov 11 at 11:46
















            • Man math is beautiful! Could you explain just the part of $x^3$ where you get $beta$ in more detail?
              – Aleksa
              Nov 11 at 11:35







            • 1




              @Aleksa I agree! :) What I wanted to add is that, for remainder and factor theorem (and polynomial division in general), I often find it useful to go back to the basic theory because it's quite obvious and often very easy when you write it out. Better not to learn too many things by rote.
              – Deepak
              Nov 11 at 11:37






            • 1




              Sorry, missed your edit. There are only two "ways" of getting the $x^3$ term of $P(x)$ on the LHS from the expression on the RHS - multiplying the $x^3$ term of $D(x)$ by the constant term of $Q(x)$ and multiplying the $x^2$ term of $D(x)$ by the $x$ term of $Q(x)$. Adding these (signed) terms up on both sides gives you a quick linear equation you can solve for $beta$.
              – Deepak
              Nov 11 at 11:39











            • Thanks for your feedback!
              – Aleksa
              Nov 11 at 11:46










            • Sure, glad to help! :)
              – Deepak
              Nov 11 at 11:46















            Man math is beautiful! Could you explain just the part of $x^3$ where you get $beta$ in more detail?
            – Aleksa
            Nov 11 at 11:35





            Man math is beautiful! Could you explain just the part of $x^3$ where you get $beta$ in more detail?
            – Aleksa
            Nov 11 at 11:35





            1




            1




            @Aleksa I agree! :) What I wanted to add is that, for remainder and factor theorem (and polynomial division in general), I often find it useful to go back to the basic theory because it's quite obvious and often very easy when you write it out. Better not to learn too many things by rote.
            – Deepak
            Nov 11 at 11:37




            @Aleksa I agree! :) What I wanted to add is that, for remainder and factor theorem (and polynomial division in general), I often find it useful to go back to the basic theory because it's quite obvious and often very easy when you write it out. Better not to learn too many things by rote.
            – Deepak
            Nov 11 at 11:37




            1




            1




            Sorry, missed your edit. There are only two "ways" of getting the $x^3$ term of $P(x)$ on the LHS from the expression on the RHS - multiplying the $x^3$ term of $D(x)$ by the constant term of $Q(x)$ and multiplying the $x^2$ term of $D(x)$ by the $x$ term of $Q(x)$. Adding these (signed) terms up on both sides gives you a quick linear equation you can solve for $beta$.
            – Deepak
            Nov 11 at 11:39





            Sorry, missed your edit. There are only two "ways" of getting the $x^3$ term of $P(x)$ on the LHS from the expression on the RHS - multiplying the $x^3$ term of $D(x)$ by the constant term of $Q(x)$ and multiplying the $x^2$ term of $D(x)$ by the $x$ term of $Q(x)$. Adding these (signed) terms up on both sides gives you a quick linear equation you can solve for $beta$.
            – Deepak
            Nov 11 at 11:39













            Thanks for your feedback!
            – Aleksa
            Nov 11 at 11:46




            Thanks for your feedback!
            – Aleksa
            Nov 11 at 11:46












            Sure, glad to help! :)
            – Deepak
            Nov 11 at 11:46




            Sure, glad to help! :)
            – Deepak
            Nov 11 at 11:46











            3














            Hint: Compute the remainder of $x^4-x^3$ when divided by $x^3+2x^2-3x+1$.






            share|cite|improve this answer




















            • I've got $ax^2+bx+c=(x^3+2x^2-3x+1)[Q(x)-(x+3)]-6x^2+8x+4$ now, what do I do from this point?
              – Aleksa
              Nov 11 at 10:59










            • The remainder was $9x^2-10x-3$, so I moved $(x^3-2x^2-3x+1)(x-3)+9x^2-10x-3$ on the other side to get that
              – Aleksa
              Nov 11 at 11:00
















            3














            Hint: Compute the remainder of $x^4-x^3$ when divided by $x^3+2x^2-3x+1$.






            share|cite|improve this answer




















            • I've got $ax^2+bx+c=(x^3+2x^2-3x+1)[Q(x)-(x+3)]-6x^2+8x+4$ now, what do I do from this point?
              – Aleksa
              Nov 11 at 10:59










            • The remainder was $9x^2-10x-3$, so I moved $(x^3-2x^2-3x+1)(x-3)+9x^2-10x-3$ on the other side to get that
              – Aleksa
              Nov 11 at 11:00














            3












            3








            3






            Hint: Compute the remainder of $x^4-x^3$ when divided by $x^3+2x^2-3x+1$.






            share|cite|improve this answer












            Hint: Compute the remainder of $x^4-x^3$ when divided by $x^3+2x^2-3x+1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 11 at 10:47









            lhf

            162k10166386




            162k10166386











            • I've got $ax^2+bx+c=(x^3+2x^2-3x+1)[Q(x)-(x+3)]-6x^2+8x+4$ now, what do I do from this point?
              – Aleksa
              Nov 11 at 10:59










            • The remainder was $9x^2-10x-3$, so I moved $(x^3-2x^2-3x+1)(x-3)+9x^2-10x-3$ on the other side to get that
              – Aleksa
              Nov 11 at 11:00

















            • I've got $ax^2+bx+c=(x^3+2x^2-3x+1)[Q(x)-(x+3)]-6x^2+8x+4$ now, what do I do from this point?
              – Aleksa
              Nov 11 at 10:59










            • The remainder was $9x^2-10x-3$, so I moved $(x^3-2x^2-3x+1)(x-3)+9x^2-10x-3$ on the other side to get that
              – Aleksa
              Nov 11 at 11:00
















            I've got $ax^2+bx+c=(x^3+2x^2-3x+1)[Q(x)-(x+3)]-6x^2+8x+4$ now, what do I do from this point?
            – Aleksa
            Nov 11 at 10:59




            I've got $ax^2+bx+c=(x^3+2x^2-3x+1)[Q(x)-(x+3)]-6x^2+8x+4$ now, what do I do from this point?
            – Aleksa
            Nov 11 at 10:59












            The remainder was $9x^2-10x-3$, so I moved $(x^3-2x^2-3x+1)(x-3)+9x^2-10x-3$ on the other side to get that
            – Aleksa
            Nov 11 at 11:00





            The remainder was $9x^2-10x-3$, so I moved $(x^3-2x^2-3x+1)(x-3)+9x^2-10x-3$ on the other side to get that
            – Aleksa
            Nov 11 at 11:00












            0














            Maybe if you go through the steps for polynomial long division you will get an answer for the remainder in terms of a and b. Then you can equate the x terms with the same order, to determine a and b.






            share|cite|improve this answer

























              0














              Maybe if you go through the steps for polynomial long division you will get an answer for the remainder in terms of a and b. Then you can equate the x terms with the same order, to determine a and b.






              share|cite|improve this answer























                0












                0








                0






                Maybe if you go through the steps for polynomial long division you will get an answer for the remainder in terms of a and b. Then you can equate the x terms with the same order, to determine a and b.






                share|cite|improve this answer












                Maybe if you go through the steps for polynomial long division you will get an answer for the remainder in terms of a and b. Then you can equate the x terms with the same order, to determine a and b.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 11 at 10:50









                KnowsNothing

                355




                355



























                    draft saved

                    draft discarded
















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid


                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.

                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid


                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.

                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2993694%2fpolynomial-division-with-remainder%23new-answer', 'question_page');

                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Use pre created SQLite database for Android project in kotlin

                    Darth Vader #20

                    Ondo