# $f$ dividing by $x + 1$ have remainder 4, when dividing with $x^2 + 1$ have remainder 2x+3. Find remainder dividing polynomial with($x+1$)($x^2+1$)

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.

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 $$r(x)=1.5x^2+2x+4.5.$$

• You didn't show how "you got" the solutions for $\,a,b,c\ \$ – Bill Dubuque Apr 12 at 21:10

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.

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