Question

Which one of the following is a polynomial?
$
  (a){\text{ }}\dfrac{{{x^2}}}{2} - \dfrac{2}{{{x^2}}} \\
  (b){\text{ }}\sqrt {2x} - 1 \\
  (c){\text{ }}{{\text{x}}^2} + \dfrac{{3{x^{\dfrac{3}{2}}}}}{{\sqrt x }} \\
  (d){\text{ }}\dfrac{{x - 1}}{{x + 1}} \\
 $

Answer 1

Hint – In this question use the basic definition of the polynomial that the power of the variable involved must not be negative, that is it has to be a non-negative integer.

Complete step-by-step solution -
General polynomial of second degree is
$a{x^2} + bx + c$
So in a polynomial the power of x must be a non-negative integer.
So check out options
Option $\left( a \right)\dfrac{{{x^2}}}{2} - \dfrac{2}{{{x^2}}}$
So above expression is also written as
$ \Rightarrow \dfrac{1}{2}{x^2} - 2{x^{ - 2}}$
So as we see that the power of x is a negative integer so it is not a polynomial.
Option $\left( b \right)\sqrt {2x} - 1$
So above expression is also written as
$ \Rightarrow \sqrt 2 {\left( x \right)^{\dfrac{1}{2}}} - 1$
So as we see that the power of x is not an integer it is in a fraction so it is not a polynomial.
Option $\left( c \right){x^2} + \dfrac{{3{x^{\dfrac{3}{2}}}}}{{\sqrt x }}$
So above expression is also written as
$ \Rightarrow {x^2} + \dfrac{{3{x^{\dfrac{3}{2}}}}}{{\sqrt x }} = {x^2} + 3\dfrac{{{x^{\dfrac{3}{2}}}}}{{{x^{\dfrac{1}{2}}}}}$
Now simplify the above expression we have,
$ \Rightarrow {x^2} + 3\dfrac{{{x^{\dfrac{3}{2}}}}}{{{x^{\dfrac{1}{2}}}}} = {x^2} + 3{x^{\dfrac{3}{2} - \dfrac{1}{2}}} = {x^2} + 3{x^{\dfrac{2}{2}}} = {x^2} + 3x$
So as we see that the power of x is positive and integer so it is a polynomial.
Option $\left( d \right)\dfrac{{x - 1}}{{x + 1}}$
As we know that a polynomial can never have a variable in the denominator so the above expression is also not a polynomial.
Hence option (c) is correct.

Note – Key point is that we solve by looking at every option and checking the degree of the variable which should be positive and integers . we simplify the given expressions.