# To find: the real zeros of polynomial function form an arithmetic sequaence f(x)=x^{4}-4x^{3}-4x^{2}+16x

To find:
The real zeros of polynomial function form an arithmetic sequaence
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{{4}}}-{4}{x}^{{{3}}}-{4}{x}^{{{2}}}+{16}{x}$$

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davonliefI
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 $$\displaystyle{4}.$$