## Precalculus (6th Edition) Blitzer

The given function $h\left( x \right)=8{{x}^{3}}-{{x}^{2}}+\frac{2}{x}$is not a polynomial.
A function is said to be a polynomial function if $g\left( x \right)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+\cdots +{{a}_{2}}{{x}^{2}}+{{a}_{1}}x+{{a}_{0}}$ , Where ${{a}_{n}},{{a}_{n-1}},\ldots ,{{a}_{2}},{{a}_{1}},{{a}_{0}}$ are any real numbers with ${{a}_{n}}\ne 0$ and n is any non-negative integer. Now, the function g(x) is called a polynomial function of degree n; its coefficient is called the leading coefficient. The given function $h\left( x \right)=8{{x}^{3}}-{{x}^{2}}+\frac{2}{x}$ , simplifies into $h\left( x \right)=8{{x}^{3}}-{{x}^{2}}+2{{x}^{-1}}$. The power of $x$ is −1, which gives a negative value for $n$ and hence the function does not satisfy all the conditions to be a polynomial. Hence, the function $h\left( x \right)=8{{x}^{3}}-{{x}^{2}}+\frac{2}{x}$ is not a polynomial.