Which of the following expressions are polynomials?

Question:

(i) 8

(ii) $\sqrt{3} x^{2}-2 x$

(iii) $1-\sqrt{5} x$

(iv) $\frac{1}{5 x^{-2}}+5 x+7$

(v) $\frac{(x-2)(x-4)}{x}$

(vi) $\frac{1}{x+1}$

(vii) $\frac{1}{7} a^{3}-\frac{2}{\sqrt{3}} a^{2}+4 a-7$

(viii) $\frac{1}{2 x}$

Solution:

(i) Polynomial, because the exponent of the variable of 8 or $8 x^{0}$ is 0 which is a whole number.

(ii) Polynomial, because the exponent of the variable of $\sqrt{3} x^{2}-2 x$ is a whole number.

(iii) Not polynomial, because the exponent of the variable of $1-\sqrt{5} x$ or $1-\sqrt{5} x^{\frac{1}{2}}$ is $\frac{1}{2}$ which is not a whole number.

(iv) Polynomial, because the exponent of the variable of $\frac{1}{5 x^{-2}}+5 x+7=\frac{1}{5} x^{2}+5 x+7$, is a whole number.

(v) Not polynomial, because the exponent of the variable of $\frac{(x-2)(x-4)}{x}=\frac{x^{2}-6 x+8}{x}$ $=x-6+8 x^{-1}$ is $-1$ which is not a whole number.

(vi) Not polynomial, as the polynomial is expressed as $a_{0}+a_{1} x+a_{2} x+\ldots a_{n} x^{n}$, where $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ are constants. Now, $f(x)=\frac{p(x)}{q(x)}$ is a rational expression where $q(x) \neq 0$, $p(x)$ and $q(x)$ are polyonimials. Hence, $\frac{1}{x+1}$ is a rational expression but not a polynomial.

(vii) Polynomial, because the exponent of the variable of $\frac{1}{7} a^{3}-\frac{2}{\sqrt{3}} a^{2}+4 a-7$ is a whole number.

(viii) Not polynomial, because the exponent of the variable of $\frac{1}{2 x}$ or $\frac{1}{2} x^{-1}$ is $-1$ which is not a whole number.