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Problem: The polynomial $f$ dividing by ($x + 1$) gives the remainder 4, and when dividing with ($x^2 + 1$) gives the remainder (2x+3). Determine the remainder when dividing the polynomial with ($x + 1$)($x^2 + 1$)?

My attempt: By Polynomial remainder theorem we know that $$f(x)=q_1(x)(x+1)+4$$ $$f(x)=q_2(x)(x^2+1)+(2x+3)$$ By putting $x=1$ we know that $f(-1)=4, f(i)=2i+3, f(-i)=-2i+3$ We want to find $r(x)$ such that: $$f(x)=(x+1)(x^2+1)q_3(x) + r(x) .$$ By applying the previous idea we know that $f(-1)=r(-1)=4$, but the same idea $r(i)=2i+3$ and $r(-i)=-2i+3$ but this is only three point and we need to determinant polynomial $r(x)$ of degree 3... Please solve without modular arithmetic.

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The hint.

Since $\deg((x+1)(x^2+1))=3,$ the degree of $r$ must be less than $3$.

Let $r(x)=ax^2+bx+c$ and take $x=-1$, $x=i$ and $x=-i$.

Now, solve the following system. $$a-b+c=4,$$ $$-a+bi+c=2i+3$$ and $$-a-bi+c=-2i+3.$$ I got $$r(x)=1.5x^2+2x+4.5.$$

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  • $\begingroup$ You didn't show how "you got" the solutions for $\,a,b,c\ \ $ $\endgroup$ – Bill Dubuque Apr 12 at 21:10
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Upon applying the law: $\,ab\bmod ac = a(b\bmod c) = $ Mod Distributive Law we obtain

$(f\!-\!(2x\!+\!3))\bmod{(x^2\!+\!1)(x\!+\!1)}\,=\, (x^2\!+\!1){\Huge[}\dfrac{\overbrace{f\!-\!(2x\!+\!3)}^{\large 4\,-\,(2(\color{#c00}{\bf -1})+3)}}{\underbrace{x^2\!+\!1}_{\large (\color{#c00}{\bf -1})^2+1\ \ \ \ }}\underbrace{\bmod{x\!+\!1}}_{\Large x\ \equiv\ \color{#c00}{\bf {-}1}}{\Huge]}=\, (x^2\!+\!1)\left[\dfrac{3}{2}\right]$

So $\, f\equiv 2x\!+\!3 + \dfrac{3}2(x^2\!+\!1).\ $ Note that we did not need to solve any equations - just evaluate.

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  • $\begingroup$ See here for further examples and discussion of the close relationship with CRT = Chinese Remainder Theorem and Lagrange interpolation. $\endgroup$ – Bill Dubuque Apr 12 at 21:20

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