UkusakazaL

2020-11-14

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

lamusesamuset

Concept:
A rational 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,$f\left(x\right)=p\frac{x}{q}\left(x\right)$ is a rational function.
Where, p(x) and g(x) are polynomial functions, with $g\left(x\right)\ne 0.$
Calculation:
The given polynomial unction form an arithmetic sequence is
$f\left(x\right)={x}^{4}-4{x}^{3}-4{x}^{2}+16x.$
Here, the constant is 0.
The above equation can be rewritten as
$f\left(x\right)=x\left({x}^{3}-4{x}^{2}-4x+16\right)$
The possibilities for $\frac{p}{q}are±1,±2,±4,and±8.$ Factoring the term $\left({x}^{3}-4{x}^{2}-4x+16\right)$, we get $\left({x}^{3}-4{x}^{2}-4x+16\right)=\left(x+2\right)\left({x}^{2}-6x+8\right)$ Factoring the term $\left({x}^{2}-6x+8\right)$, we get $\left({x}^{2}-6x+8\right)=\left(x-2\right)\left(x-4\right)$ Combining all the terms, we get $\left({x}^{3}-4{x}^{2}-4x+16\right)=\left(x+2\right)\left(x-2\right)\left(x-4\right)$
$f\left(x\right)=x\left({x}^{3}-4{x}^{2}-4x+16\right)=x\left(x+2\right)\left(x-2\right)\left(x-4\right)$ Thus, the real zeros are -2, 0, 2, and 4

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