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What is the remainder when: $$x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41$$ is divided by: $$x^2-8x+12$$

$x^2-8x+12$ $\leftrightarrow (x-2)(x-6)$

This gives me the polynomial: $(x-2)(x-6)k(x) + r(x)$, where $r(x)$ is the remainder.

The expression gives me the remainder when $x=2$ or $x = 6$

By replacing $x$ with $2$ into the original polynomial you get: $$x^{100} - 4x^{100}+3x^{100}-3x^{10}+12x^{10}-9x^{10}+3x^{2}-29x + 41 = 3x^2-29x + 41$$

By plugging in $2$ I get: $12-58 + 41 = -5$.

Hence $r(x)$ should be: $-5$.

This is apparently the wrong answer. What did I do wrong?

Thank you kindly for your help!

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    $\begingroup$ "By entering $2$ into the original polynomial you get:" - don't you mean to replace $x$ with $2$? $\endgroup$
    – Git Gud
    Feb 24, 2013 at 16:56
  • $\begingroup$ Yes that is right, sorry for my mistake. $\endgroup$ Feb 24, 2013 at 16:57
  • $\begingroup$ "Hence $r(x)=-5$". What you can conclude is that $r(2)=-5$. $\endgroup$
    – Git Gud
    Feb 24, 2013 at 16:58
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    $\begingroup$ You could start by noticing that the first three terms can be ignored. $\endgroup$ Feb 24, 2013 at 17:08
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    $\begingroup$ +1 to OP for clearly writing out his problem and indicating his steps. $\endgroup$ Feb 24, 2013 at 17:36

3 Answers 3

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First note that $r(x)=ax+b$ for some $a,b\in\mathbb Z$
For $x=2$ you get $r(2)=-5$ (not $r(x)=-5$) $ \Rightarrow 2a+b=-5$. Do the same for $x=6$ to find r(6).Then find $a,b$.

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Hint $\ $ You are very close to the solution. By the polynomial division algorithm

$$\begin{eqnarray} &&\rm f(x) &\: =\: &\rm a + bx + (x-2)(x-6)\, g(x)\\ \Rightarrow\ &&\rm f(2) &\:=\:&\rm a + 2b \\ \Rightarrow\ &&\rm f(6) &\:=\:&\rm a + 6b \end{eqnarray}$$

Solving the above two linear equations for $\rm\,a,b\,$ determines the remainder. Note that the system has a unique solution since the determinant $\ne 0,\,$ being the difference of the distinct evaluation points.

Remark $\ $ The same method extends from two points to $\rm\,n\,$ points, i.e. we can uniquely interpolate the degree $\rm < n\,$ remainder through any $\rm\,n\,$ distinct points because a polynomial of degree $\rm < n\,$ over a field is determined uniquely by $\rm\,n\,$ values. Indeed, the above system of equations has Vandermonde determinant equal to the product of the differences of the evaluation points, so it is be nonzero if the roots are distinct. See here for an example.

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It is worth noting that $$ \begin{multline} x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41\\ =x^{98}(x^2-8x+12)-3x^8(x^2-8x+12)+3(x^2-8x+12)-5x+5 \end{multline} $$ so that the remainder is $-5x+5$.

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