Step 1

The might know that

PSKa+I=b+I\Leftrightarrow a-b\in IZK

So

\(\displaystyle{\left({x}^{{{3}}}+{x}^{{{2}}}\right)}-{1}={\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}\in{I}\)

So

\(\displaystyle{\left({x}^{{{3}}}+{x}^{{{2}}}\right)}+{I}={1}+{I}\)

also any element in \(\displaystyle{\frac{{{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}}}{{{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}}}}\) is of the form \(\displaystyle{\left({a}{x}^{{{2}}}+{b}{x}+{c}\right)}+{I}\). There are 2 choices for each a, b, c. So there are 8 distinct elements in \(\displaystyle{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}\) by a cubic polynomial, you get a remainder which is a quadratic polynomial of the form \(\displaystyle{a}{x}^{{{2}}}+{b}{x}+{c}\)

So if

\(\displaystyle{p}{\left({x}\right)}\in{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}{]}\) then \(\displaystyle{p}{\left({x}\right)}={q}{\left({x}\right)}{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}+{r}{\left({x}\right)}\)

where \(\displaystyle{r}{\left({x}\right)}={a}{x}^{{{2}}}+{b}{x}+{c}\). For some \(\displaystyle{a},\ {b},\ {c}\in{\mathbb{{{Z}}}}_{{{2}}}\)

So

\(\displaystyle{p}{\left({x}\right)}+{I}={a}{x}^{{{2}}}+{b}{x}+{c}+{I}\)

as

\(\displaystyle{q}{\left({x}\right)}{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}\in{\left\langle{x}^{{{3}}}+{x}^{{{2}}}+{1}\right\rangle}\)

The might know that

PSKa+I=b+I\Leftrightarrow a-b\in IZK

So

\(\displaystyle{\left({x}^{{{3}}}+{x}^{{{2}}}\right)}-{1}={\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}\in{I}\)

So

\(\displaystyle{\left({x}^{{{3}}}+{x}^{{{2}}}\right)}+{I}={1}+{I}\)

also any element in \(\displaystyle{\frac{{{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}}}{{{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}}}}\) is of the form \(\displaystyle{\left({a}{x}^{{{2}}}+{b}{x}+{c}\right)}+{I}\). There are 2 choices for each a, b, c. So there are 8 distinct elements in \(\displaystyle{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}\) by a cubic polynomial, you get a remainder which is a quadratic polynomial of the form \(\displaystyle{a}{x}^{{{2}}}+{b}{x}+{c}\)

So if

\(\displaystyle{p}{\left({x}\right)}\in{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}{]}\) then \(\displaystyle{p}{\left({x}\right)}={q}{\left({x}\right)}{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}+{r}{\left({x}\right)}\)

where \(\displaystyle{r}{\left({x}\right)}={a}{x}^{{{2}}}+{b}{x}+{c}\). For some \(\displaystyle{a},\ {b},\ {c}\in{\mathbb{{{Z}}}}_{{{2}}}\)

So

\(\displaystyle{p}{\left({x}\right)}+{I}={a}{x}^{{{2}}}+{b}{x}+{c}+{I}\)

as

\(\displaystyle{q}{\left({x}\right)}{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}\in{\left\langle{x}^{{{3}}}+{x}^{{{2}}}+{1}\right\rangle}\)