A ration expression is a fraction that is quotient of two polynomials. A rational function is defined by the two polynomial functions.

A function f of the form \(\displaystyle{f{{\left({x}\right)}}}={\frac{{{p}{\left({x}\right)}}}{{{q}{\left({x}\right)}}}}\) is a rational function.

Where, \(\displaystyle{p}{\left({x}\right)}\) and \(\displaystyle{q}{\left({x}\right)}\) are polynomial functions, with \(\displaystyle{q}{\left({x}\right)}\ne{q}{0}.\)

Calculation:

The given polinomial function form an arithmetic sequence is

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{4}}}-{4}{x}^{{{3}}}-{4}{x}^{{{2}}}+{16}{x}\)

Here, the constant is 0.

The above equatioon can be rewritten as

\(\displaystyle{f{{\left({x}\right)}}}={x}{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)}\)

The possiblities for \(\displaystyle{\frac{{{p}}}{{{q}}}}\) are \(\displaystyle\pm{1},\ \pm{2},\ \pm{4}\) and \(\displaystyle\pm{8}\)

Factoring the term \(\displaystyle{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)},\) we get

\(\displaystyle{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)}={\left({x}+{2}\right)}{\left({x}^{{{2}}}-{6}{x}+{8}\right)}\)

Factoring the term \(\displaystyle{\left({x}^{{{2}}}-{6}{x}+{8}\right)},\) we get

\(\displaystyle{\left({x}^{{{2}}}-{6}{x}+{8}\right)}={\left({x}-{2}\right)}{\left({x}-{4}\right)}\)

Combining all the terms, we get

\(\displaystyle{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)}={\left({x}+{2}\right)}{\left({x}-{2}\right)}{\left({x}-{4}\right)}\)

So, \(\displaystyle{f{{\left({x}\right)}}}={x}{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)}={x}{\left({x}+{2}\right)}{\left({x}-{2}\right)}{\left({x}-{4}\right)}\)

Thus, the real zeros are \(\displaystyle-{2},\ {0},\ {2},\) and \(\displaystyle{4}.\)

A function f of the form \(\displaystyle{f{{\left({x}\right)}}}={\frac{{{p}{\left({x}\right)}}}{{{q}{\left({x}\right)}}}}\) is a rational function.

Where, \(\displaystyle{p}{\left({x}\right)}\) and \(\displaystyle{q}{\left({x}\right)}\) are polynomial functions, with \(\displaystyle{q}{\left({x}\right)}\ne{q}{0}.\)

Calculation:

The given polinomial function form an arithmetic sequence is

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{4}}}-{4}{x}^{{{3}}}-{4}{x}^{{{2}}}+{16}{x}\)

Here, the constant is 0.

The above equatioon can be rewritten as

\(\displaystyle{f{{\left({x}\right)}}}={x}{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)}\)

The possiblities for \(\displaystyle{\frac{{{p}}}{{{q}}}}\) are \(\displaystyle\pm{1},\ \pm{2},\ \pm{4}\) and \(\displaystyle\pm{8}\)

Factoring the term \(\displaystyle{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)},\) we get

\(\displaystyle{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)}={\left({x}+{2}\right)}{\left({x}^{{{2}}}-{6}{x}+{8}\right)}\)

Factoring the term \(\displaystyle{\left({x}^{{{2}}}-{6}{x}+{8}\right)},\) we get

\(\displaystyle{\left({x}^{{{2}}}-{6}{x}+{8}\right)}={\left({x}-{2}\right)}{\left({x}-{4}\right)}\)

Combining all the terms, we get

\(\displaystyle{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)}={\left({x}+{2}\right)}{\left({x}-{2}\right)}{\left({x}-{4}\right)}\)

So, \(\displaystyle{f{{\left({x}\right)}}}={x}{\left({x}^{{{3}}}-{4}{x}^{{{2}}}-{4}{x}+{16}\right)}={x}{\left({x}+{2}\right)}{\left({x}-{2}\right)}{\left({x}-{4}\right)}\)

Thus, the real zeros are \(\displaystyle-{2},\ {0},\ {2},\) and \(\displaystyle{4}.\)