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