Consider the following polynomials over Z8 where a is written for [a] in Z8:- f(x)=2x3+7x+4,g(x)=4x2+4x+6,h(x)=6x2+3....

Step 1
Given :
The given polynomials over ${\displaystyle Z_8}$ are
${\displaystyle f\left(x\right)=2x^3+7x+4,g\left(x\right)=4x^2+4x+6}$
Compute f(x)+g(x) over ${\displaystyle Z_8}$
Step 2
${\displaystyle f\left(x\right)+g\left(x\right)=(2x^3+7x+4)+(4x^2+4x+6)}$
${\displaystyle =8x^5+8x^4+12x^3+28x^3+28x^2+42x+16x^2+16x+24}$
${\displaystyle =0+0+4x^3+4x^3+4x^2+2x+0+0+0}$
${\displaystyle =0+0+0+4x^2+2x+0+0+0}$
${\displaystyle =4x^2+2x}$

We need to find f(x)+g(x) of the following polynomials ${\displaystyle f\left(x\right)=2x^3+7x+4,g\left(x\right)=4x^2+4x+6}$, with all coefficients in ${\displaystyle Z_8}$.
If ${\displaystyle f\left(x\right)=\sum _{i=0}^n{a}_i{x}^i}$ and ${\displaystyle g\left(x\right)=\sum _{i=0}^m{b}_i{x}^i}$ in R[x], we define addition in R[x] by ${\displaystyle f\left(x\right)+g\left(x\right)=\sum _{i=0}^k(a_i+b_i)x^i}$,
where k is the larger of the tow integers n, m.
Thus, we have
${\displaystyle f\left(x\right)+g\left(x\right)=(2x^3+7x+4)+(4x^2+4x+6)}$
${\displaystyle =(2+0)x^3+(0+4)x^2+(7+4)x+(4+6)}$
${\displaystyle =2x^3+4x^2+11x+10}$
Now, we have
${\displaystyle {\left[11\right]}_8={\left[3\right]}_8}$ and ${\displaystyle {\left[10\right]}_8={\left[2\right]}_8}$.
Thus, we get
${\displaystyle f\left(x\right)+g\left(x\right)=2x^3+4x^2+3x+2}$.
Hence, ${\displaystyle f\left(x\right)+g\left(x\right)=2x^3+4x^2+3x+2}$.

Given, $f(x)=2x^3+7x+4,g(x)=4x^2+4x+6$ Step 2 Now, $f(x)+g(x)=(2x^3+7x+4)+(4x^2+4x+6)$ $=2x^3+4x^2+(7x+4x)+(4+6)$ $=2x^3+4x^2+11x+10$ Now, $[11{]}_8=[3{]}_8$ & $[10{]}_8=[2{]}_8$ Therefore, $f(x)+g(x)=2x^3+4x^2+3x+2$