$(3x^2+2x+1)(4x^3+x^2+5x+1)=5x^5+4x^4+1\equiv 1 \pmod {x^4}$
Expanding the first part, I get $12x^5+11x^4+21x^3+14x^2+7x+1$. However, I do not understand how to get from the above statement to $5x^5+4x^4+1\equiv 1 \pmod{x^4}$.
Another example: $(2x+1)(2x^2+x+1)= 4x^3+4x^2+3x+1 \equiv 1 \pmod {x^2}$.
Any help with understanding this concept would be greatly appreciated. Thanks!
Additional Note: (Not sure if this helps at all)
The original question involves Newton's Iteration. Let $a=5x^5 + 4x^4 + 3x^3 + 2x^2 + x$ and $b = x^2 + 2x + 3$ be polynomials. Then the inverse of the reversal of b, multiplied by the reversal of b, must $\equiv 1 mod x^4$.
The first congruence is true $\iff 7\,(3x^3\!+2x^2\!+x)\equiv 0 \pmod{x^4}.\, $ To further deduce that it's $\equiv 0$ you need further hypothesis, $ $ e.g. $\ 7\equiv 0.\,$ Lacking such the claim is false.
Similary, the second congruence holds $\iff 3x\equiv 0\pmod{x^2},\ $ e.g. if $\ 3\equiv 0.$