I Solve the problem involving Rings

[chwala]

TL;DR Summary

Kindly see attached (reference is highlighted part)

Its a bit clear; i can follow just to pick another polynomial say

$(x+1)^3$ are we then going to have $(2x-2)+ x+3$?

or it has to be a polynomial with

$x^2+1$ being evident? cheers...

 

[fresh_42]

Let's see ...
\begin{align*}
(x+1)^3&=(x+1)^2\cdot (x+1)=2x\cdot (x+1)=2x^2+2x=2\cdot (x^2+1) -2+2x=2x-2=2\cdot (x-1)\\
&\text{crosscheck}\\
(x+1)^3&=x^3+3x^2+3x+1=x\cdot (-1)+3\cdot (-1)+3x+1=2x-2=2\cdot(x-1)
\end{align*}
We identify $x^2=-1.$

 

[chwala]

...I was getting a bit lost on this line:

$$2x^2+2x=2\cdot (x^2+1) -2+2x=2x-2=2\cdot (x-1)$$

...but i now i got it by letting,

$x^2+x=x^2+1+m$

therefore

$m=x-1$

other steps follow well.

 

[chwala]

...which means that

$(x+1)^4=(x+1)^2 (x+1)^2 =2x ⋅2x=4x^2=4(-1)=-4$

Cross check:

$(x+1)^4=x^4+4x^3+6x^2+4x+1=1-4x-6+4x+1=-4$

 

[chwala]

Lastly on this just to get the drift of things, if we have a polynomial say;

$x^3+4$

then this becomes,

$x(x^2)+4=x(-1)+4=4-x$

...

 

[fresh_42]

Lastly on this just to get the drift of things, if we have a polynomial say;

$x^3+4$

then this becomes,

$x(x^2)+4=x(-1)+4=4-x$

...

Right. The ring $\mathbb{R}[x]$ is an integral domain, the ideal $\langle x^2+1 \rangle$ is prime and maximal. This makes $\mathbb{R}[x]/\langle x^2+1 \rangle$ a field, the complex numbers. So $x= \mathrm{i}.$

Here we have $x^3+4= \mathrm{i}^3+4=-\mathrm{i} +4 =4-x.$