Question

Which of the following is a polynomial of x?
A) $$x+\dfrac{1}{x}$$
B) $$x^{2}+\sqrt{x}$$
C) $$x+\sqrt{2} x^{2}+1$$
D) $$\sqrt{3x} +1$$

Answer 1

Hint:
 In this question, it is given that we have to find among the above options which is a polynomial of x. So to find the solution we need to know that, in each term of an algebraic expression if the power of ‘x’ (variable)is Natural number (N) then the expression is called a polynomial.

Complete step by step answer:
First we check for option A.
 $$x+\dfrac{1}{x}$$
$$=x+x^{-1}$$ [Since $$\dfrac{1}{a} =a^{-1}$$]
Here the power of x is (-1), which is not a Natural number.
Therefore the given algebraic expression is not a polynomial.

Option B,
$$x^{2}+\sqrt{x}$$
=$$x^{2}+x^{\dfrac{1}{2} }$$
Here in the second term the power of x is $$\dfrac{1}{2}$$ which is not a Natural number.
Therefore it is not a polynomial.

Option C,
$$x+\sqrt{2} x^{2}+1$$
Here the power of x in each term is Natural number, i.e in the first term power is 1, in the second term it is 2 and in the third term the power of x is zero.
So the above algebraic expression is a polynomial.

Option D,
$$\sqrt{3x} +1$$
=$$\sqrt{3} \sqrt{x} +1$$ [Since $$\sqrt{ab} =\sqrt{a} \times \sqrt{b}$$]
=$$\sqrt{3} x^{\dfrac{1}{2} }+1$$
In the first term the power of x is $$\dfrac{1}{2}$$ therefore the expression is not a polynomial.
So therefore we can say that option C is a polynomial.

Note:
Polynomials are the algebraic expressions that consist of variables and coefficients. Variables are also sometimes called indeterminates. We can perform the arithmetic operations such as addition, subtraction, multiplication and also positive integer exponents for polynomial expressions but not division by variable. An example of a polynomial with one variable is $$x^{2}+2x-10$$. In this example, there are three terms: $$x^{2}$$, 2x and -10.