Question

Which of the following expressions are a polynomial?
A. $x+\dfrac{1}{x}$
B. $\sqrt{x}+x+{{x}^{2}}$
C. $\sqrt{2}x+{{x}^{3}}+3{{x}^{2}}$
D. ${{x}^{2}}-{{x}^{-2}}-2$

Answer 1

Hint: We use the condition of an algebraic expression to be called a polynomial. It gives that all the exponents in the algebraic expression must be non-negative integers. We transform all the given expressions in the exponent form of the variable. We find if any negative exponents or fraction is there and if not, we call it a polynomial.

Complete step by step answer:
The conditions of an algebraic expression to be called a polynomial is that all the exponents in the algebraic expression must be non-negative integers. This means if the exponent or power of a variable is n then $n\ge 0,n\in \mathbb{Z}$.
Now we try to express the given expressions in the form of an exponent and find out if that is a polynomial or not.
For $x+\dfrac{1}{x}$, we convert it in the form of $x+\dfrac{1}{x}\to {{x}^{1}}+{{x}^{-1}}$. We can see that there is an exponent of -1 which is negative. So, it can’t be a polynomial.
For $\sqrt{x}+x+{{x}^{2}}$, we convert it in the form of $\sqrt{x}+x+{{x}^{2}}\to {{x}^{\dfrac{1}{2}}}+x+{{x}^{2}}$. We can see that there is an exponent of $\dfrac{1}{2}$ which is a fraction. So, it can’t be a polynomial.
For $\sqrt{2}x+{{x}^{3}}+3{{x}^{2}}$, we convert it in the form of $\sqrt{2}x+{{x}^{3}}+3{{x}^{2}}\to \sqrt{2}.{{x}^{1}}+{{x}^{3}}+3{{x}^{2}}$. We can see that all the exponents are non-negative integers. So, it is a polynomial.
For ${{x}^{2}}-{{x}^{-2}}-2$, we convert it in the form of ${{x}^{2}}-{{x}^{-2}}-2\to {{x}^{2}}-{{x}^{-2}}-2{{x}^{0}}$. We can see that there is an exponent of -2 which is negative. So, it can’t be a polynomial.

So, the correct answer is “Option C”.

Note: We do not need to worry about any constants or its exponent or power value. Constants can be anything. We only need to care about the variables. In $\sqrt{2}x+{{x}^{3}}+3{{x}^{2}}$, there was a constant of $\sqrt{2}$ but it was still a polynomial.