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I took an exam today and there's a problem stuck in my head; I still can't figure out yet.

Here's the question (just the concept as I can't remember precisely).

An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the number) and when this polynomial is divided by $x^2 $, it leaves $2x + 4$ (again, not sure about the number). From the given conditions, if this polynomial is divided by $(x-1)x^2$, what would be the remainder?

The solution as far as I figured out is this:

first, from the division of $(x-1)^2$, I got that $f(1) = 3$ in the same way from division of $x^2$, I got $f(0) = 4.$

I can write the polynomial as follows:

$f(x) = (x-1)(x)(x) g(x) + ax^2 +bx +c$

$ax^2 + bx + c$ is the remainder. And to find $a,b,c$, I can use the conditions above, so I got $c = 4$ by substituting $x = 0,$ and I got $a+b+4 = 3$ by substituting $x = 1.$

This leaves $a + b = -1,$ and I can't figure out how to continue; please help.

Edit : I made a mistake $f(1)$ should be equal to $4$ and $a+b+c = 4$

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  • $\begingroup$ Same as in the variant a couple days ago. $\endgroup$ – Bill Dubuque Jun 23 at 14:28
  • $\begingroup$ There is a mistake in your approach, you should get $f(1)=1+3=4$ $\endgroup$ – Julian Mejia Jun 23 at 15:21
  • $\begingroup$ Calculating the two remainders of $f(x) = x^2(x-1)^2h(x) + px^3+qx^2 +rx +s$ and setting them equal to the given remainders gives 4 equations in $p$, $q$, $r$ and $s$. $\endgroup$ – random Jun 24 at 14:01
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We have

$$f(x)=(x-1)^2q_1(x)+x+3$$

$$f'(x)=2(x-1)q_1(x)+(x-1)^2q_1'(x)+1$$

Where for $x=1$ we have $f(1)=4$ and $f'(1)=1$

Then from

$$f(x)=x^2q_2(x)+2x+4$$ $$f'(x)=2xq_2(x)+x^2q_2'(x)+2$$

Where for $x=0$ we have $f(0)=4$ and $f'(0)=2$

Now from $$f(x)=x^2(x-1)q_3(x)+ax^2+bx+c$$ and

$$f'(x)=x^2q_3(x)+2x(x-1)q_3(x)+(x-1)x^2q_3'(x)+2ax+b$$

When we substitute the values $x=0$ and $x=1$ in $f$ and $f'$ we get

$f(0)=c=4$ and $f(1)=a+b+4=4$ $$a+b=0$$

$f'(0)=b=2$ from this we have $a=-2$. Thus the remainder is $r(x)=-2x^2+2x+4$

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I am going to follow your approach. We have $$f(x)=(x-1)^2p(x)+x+3$$ So, $f(1)=4$. We also have that $$f(x)=x^2q(x)+2x+4$$

And we want to find $a,b,c$ in $$f(x)=(x-1)x^2g(x)+ax^2+bx+c$$ Plugin $x=1$ you get $4=a+b+c$.

Plugin $x=0$ doesn't give us enough information since this only gives us the remainder after dividing by $x$(instead of $x^2$).

So, instead of plugin, we look $f$ mod $x^2$. We have $$f(x)=x^2[(x-1)g(x)+a]+(bx+c)$$ Since we are given that the remainder of dividing $f(x)$ by $x^2$ is $2x+4$, we conclude $bx+c=2x+4$, i.e. $b=2$, $c=4$. And therefore, $a=-2$.

We conclude that the reminder of $f$ dividing by $ (x-1)x^2$ is $-2x^2+2x+4$.

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  • $\begingroup$ Thanks a lot ! but i'm not quite understand about not plugging in x = 0 ,can you explain more? $\endgroup$ – TAP TAP TAP Jun 23 at 15:45
  • $\begingroup$ @TAP, What I am saying is that plugging in $x=0$ is the same as saying that the remainder of dividing $f$ by $x$ is $c$ and that's why you get $c=4$. Instead, I am saying that the remainder of dividing $f$ by $x^2$ is $bx+c$ and that's why you get $bx+c=2x+4$. You get more information by doing this. So, I am not saying is wrong to plugin, but you lose information. $\endgroup$ – Julian Mejia Jun 23 at 17:11
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$\ f = \color{#0a0}{3+x + q\cdot (x\!-\!1)^2}\ $ by hypothesis, and also by hypothesis we have

$\ f = 4\!+\!2x + \color{#c00}g\cdot x^2.\, $ Put $\,\color{#c00}{g = a} + \color{#89c}{(x\!-\!1)\,h}\,\ $ ($=$ division of $\,g\,$ by $x\!-\!1)\ $ so

$\bbox[6px,border:1px solid #c00]{ f = 4\!+\!2x + \color{#c00}a\cdot x^2 + x^2\color{#89c}{(x\!-\!1)\,h}}\, =\, \color{#0a0}{3\!+\!x + q\cdot (x\!-\!1)^2} $

Eval'ed at $\,x=1\:\Rightarrow\: 4+2+\color{#c00}a+\color{#89c}0\: =\, \color{#0a0}{3\!+\!1 +0}\ $ so $\ \bbox[6px,border:1px solid #c00]{\color{#c00}{a = -2}}\ \ $ QED

Remark $ $ If you know (Easy) CRT then it immediately yields the general result as below

$\begin{align}&f\equiv \color{#c00}a\!\!\!\pmod{\!\color{#c00}g}\\ &f\equiv\color{#0a0}b\!\!\!\pmod{\!x\!-\!1}\end{align}\!\!\!\!\iff\!\! f \equiv a\! +\! g\left[ \dfrac{b\!-\!a}g\bmod x\!-\!1\right]\equiv \color{}a\! +\! \left[ \color{#c00}{\dfrac{\color{#0a0}{b(1)}\!-\!a(1)}{g(1)}}\right] g\ \pmod{(x\!-\!1)g}$

$\begin{align}&f\,\equiv\, \color{#c00}{4+2x}\!\pmod{\!x^2}\\ &f\,\equiv\,\color{#0a0}{3\,+\,x} \pmod{\!x-1}\end{align}\ \ \ \iff\ \ \ \ f\ \equiv\ \color{}{4 + 2x} \ +\ \ \underbrace{\left[\color{#c00}{\dfrac{\color{#0a0}{3\!+\!1}\!-\!(4\!+\!2)}{1^2} }\right]}_{\Large -2\ \ \ }x^2 \pmod{(x\!-\!1)x^2}$

The computation is so easy because we chose $\,x\!-\!1\,$ (vs. $x^2)\,$ as the modulus in the formula, which simplifies mod arithmetic since $\,f(x)\bmod x\!-\!1 = \color{#c00}{f(1)}\,$ by the Polynomial Remainder Theorem. Generally CRT computations are simpler when we solve last the congruences with least moduli.

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    $\begingroup$ If anything above is not clear then please feel welcome to ask questions and I will be happy to elaborate. $\endgroup$ – Bill Dubuque Jun 23 at 14:43
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General method, without the formal derivatives.

Suppose $$ f(x) = g(x) (x-1)^2 + (x+3)= h(x) x^2 + (2x + 4). $$ Then $$ (x-1)^2 f(x) = (x-1)^2 x^2 h(x) + (x-1)^2 (2x+4) \tag 1 $$ and $$ x^2 f(x) = x^2 (x-1)^2 g(x) + x^2(x+3). \tag 2 $$ Now do the Euclidean algorithm to $x^2, (x-1)^2$: \begin{align*} x^2 &= (x-1)^2 + (2x-1), \\ (x-1)^2 &= \frac 14 (2x-1)(2x - 3) + \frac 14, \end{align*} therefore $$ 1 = 4(x-1)^2 - (2x - 1) (2x-3) = 4(x-1)^2 - (x^2 - (x-1)^2) (2x-3) = (x-1)^2 (4 + 2x-3) - x^2 (2x - 3) = \color{blue}{(x-1)^2 (2x+1) - x^2 (2x - 3)}. $$ Thus $(2x+1) \cdot \mathrm {Eq}(1) - (2x - 3) \cdot \mathrm {Eq} (2)$ yields $$ f(x) = (x-1)^2 x^2 F(x) + (2x+4) (2x+1)(x-1)^2 - (x+3)(2x-3)x^2. $$ Since $$ (2x+4) (2x+1)(x-1)^2 - (x+3)(2x-3)x^2 = (4x^2 + 10x + 4)(x-1)^2 - (2x^2 + 3x -9)x^2 = (4x^3 +6x^2 -6x - 4)(x-1) - ((2x+5)(x-1) -4)x^2 = x^2 (x-1) G(x) + ((-6x -4)(x-1) + 4x^2) = x^2(x-1)G(x) + (\color{red}{-2x^2 +2x +4}), $$ we have $$ f(x) = x(x-1)^2 (G(x)+ xF(x)) + (\color{red}{-2x^2 +2x +4}), $$ which means the remainder is $$ \color{red}{-2x^2 +2x +4}. $$

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  • $\begingroup$ where did F(x) come from ? $\endgroup$ – TAP TAP TAP Jun 23 at 16:45
  • $\begingroup$ @TAPTAPTAP To be exact, $$F(x) = (2x+1) h(x) - (2x-3)g(x). $$ $\endgroup$ – xbh Jun 23 at 16:47
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Adapting the Extended Euclidean Algorithm, as implemented in this answer, to polynomial division and rotating to go down the page rather than across: $$ \begin{array}{c|cc|c} \color{#C00}{x^2}&\color{#C00}{1}&0\\ \color{#090}{x^2-2x+1}&0&\color{#090}{1}\\ 2x-1&1&-1&1\\ \color{#C90}{\frac14}&\color{#C00}{-\frac12x+\frac34}&\color{#090}{\frac12x+\frac14}&\frac12x-\frac34\\ 0&4x^2-8x+4&-4x^2&8x-4 \end{array}\tag1 $$ which says $$ \overbrace{(2x+1)}^{4\left(\frac12x+\frac14\right)}(x-1)^2+\overbrace{(-2x+3)}^{4\left(-\frac12x+\frac34\right)}x^2=\overbrace{\ \quad1\ \quad}^{4\cdot\frac14}\tag2 $$ Therefore $$ (2x+1)(x-1)^2\equiv\left\{\begin{align}1&\pmod{x^2}\\0&\pmod{(x-1)^2}\end{align}\right.\tag3 $$ and $$ (-2x+3)x^2\equiv\left\{\begin{align}0&\pmod{x^2}\\1&\pmod{(x-1)^2}\end{align}\right.\tag4 $$ Thus, mod $x^2(x-1)^2$, the polynomial is $$ \begin{align} &(2x+4)\overbrace{(2x+1)(x-1)^2}^{(3)}+(x+3)\overbrace{(-2x+3)x^2}^{(4)}\\ &=2x^4-x^3-3x^2+2x+4\\ &\equiv3x^3-5x^2+2x+4&&\bmod x^2(x-1)^2\\ &\equiv-2x^2+2x+4&&\bmod x^2(x-1)\tag5 \end{align} $$ which we can say since $\left.x^2(x-1)\,\middle|\,x^2(x-1)^2\right.$.

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$f(x)\equiv{x+3}\pmod{(x-1)^2}\wedge f(x)\equiv{2x+4}\pmod{x^2} \overset{CRT}{\implies} f(x)\equiv{3x^3 - 5x^2 + 2x + 4}\pmod{(x-1)^2x^2}$

? chinese(Mod(x+3,(x-1)^2),Mod(2*x+4,x^2))
%1 = Mod(3*x^3 - 5*x^2 + 2*x + 4, x^4 - 2*x^3 + x^2)

Then $f(x)\equiv{-2x^2 + 2x + 4}\pmod{(x-1)x^2}$

? Mod(3*x^3 - 5*x^2 + 2*x + 4,(x-1)*x^2)
%2 = Mod(-2*x^2 + 2*x + 4, x^3 - x^2)
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