Let $f(x) =7x^{32}+5x^{22}+3x^{12}+x^2$.
(i) Then find the remainder when $f(x)$ is divided by $[x^2+1]$.
(ii) Also find the remainder when $xf(x)$ is divided by $[x^2+1]$.
Given both the remainders will be of the form $4(ax+b)$.
The 'polynomial long division method' can be applied, but that won't be logical enough, because its too long a process.
Can you tell me any other shorter & easier process?
Thanks.
For the first case, note that only even powers of $x$ are involved, so you can substitute $y=x^2$. In this case you should find the remainder of $7y^{16}+5y^{11} + 3y^6 + y$ divided by $y+1$, i.e. $$7(-1)^{16}+5(-1)^{11} + 3(-1)^6 + (-1) = 4$$
So you have $f(x) = q(x)(x^2+1)+4$, where $q(x)$ is some polynomial.
From this follows that $xf(x) = xq(x)(x^2+1) +4x$, so the second remainder is $4x$.
(i) Note that $f(i)=7-5+3-1=4$, so the remainder is $4$.
(ii) Note that $if(i)=7i-5i+3i-i=4i$, so the remainder is $4x$.
This method uses the isomorphism $\dfrac{\mathbb{Z}[x]}{(x^2+1)}\simeq\mathbb{Z}[i]$, with $\varphi(x)=i$.
Hint $\ {\rm mod}\,\ x^2\!+1\!:\,\ \color{#c00}{x^2\equiv -1}\,\Rightarrow\, f(\color{#c00}{x^2})\equiv f(\color{#c00}{-1}),\ $ $\,xf(\color{#c00}{x^2})\equiv xf(\color{#c00}{-1}),\ $ for $\ f\in\Bbb Z[x]$
