2022-01-14

Consider the following polynomials over ${Z}_{8}$ where a is written for [a] in ${Z}_{8}$:-
$f\left(x\right)=2{x}^{3}+7x+4,g\left(x\right)=4{x}^{2}+4x+6,h\left(x\right)=6{x}^{2}+3$.
Find each of the following polynomials with all coefficients in ${Z}_{8}$,
f(x)+g(x)

puhnut1m

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

scomparve5j

We need to find f(x)+g(x) of the following polynomials $f\left(x\right)=2{x}^{3}+7x+4,g\left(x\right)=4{x}^{2}+4x+6$, with all coefficients in ${Z}_{8}$.
If $f\left(x\right)=\sum _{i=0}^{n}{a}_{i}{x}^{i}$ and $g\left(x\right)=\sum _{i=0}^{m}{b}_{i}{x}^{i}$ in R[x], we define addition in R[x] by $f\left(x\right)+g\left(x\right)=\sum _{i=0}^{k}\left({a}_{i}+{b}_{i}\right){x}^{i}$,
where k is the larger of the tow integers n, m.
Thus, we have
$f\left(x\right)+g\left(x\right)=\left(2{x}^{3}+7x+4\right)+\left(4{x}^{2}+4x+6\right)$
$=\left(2+0\right){x}^{3}+\left(0+4\right){x}^{2}+\left(7+4\right)x+\left(4+6\right)$
$=2{x}^{3}+4{x}^{2}+11x+10$
Now, we have
${\left[11\right]}_{8}={\left[3\right]}_{8}$ and ${\left[10\right]}_{8}={\left[2\right]}_{8}$.
Thus, we get
$f\left(x\right)+g\left(x\right)=2{x}^{3}+4{x}^{2}+3x+2$.
Hence, $f\left(x\right)+g\left(x\right)=2{x}^{3}+4{x}^{2}+3x+2$.

alenahelenash

Given, $f\left(x\right)=2{x}^{3}+7x+4,g\left(x\right)=4{x}^{2}+4x+6$ Step 2 Now, $f\left(x\right)+g\left(x\right)=\left(2{x}^{3}+7x+4\right)+\left(4{x}^{2}+4x+6\right)$ $=2{x}^{3}+4{x}^{2}+\left(7x+4x\right)+\left(4+6\right)$ $=2{x}^{3}+4{x}^{2}+11x+10$ Now, $\left[11{\right]}_{8}=\left[3{\right]}_{8}$ & $\left[10{\right]}_{8}=\left[2{\right]}_{8}$ Therefore, $f\left(x\right)+g\left(x\right)=2{x}^{3}+4{x}^{2}+3x+2$