
Identify polynomials in the following:
(a)$f\left( x \right)=4{{x}^{3}}-{{x}^{2}}-3x+7$
(b)$g\left( x \right)=2{{x}^{3}}-3{{x}^{2}}+\sqrt{x}-1$
(c)$p\left( x \right)=\dfrac{2{{x}^{2}}}{3}-\dfrac{7x}{4}+9$
(d)$q\left( x \right)=2{{x}^{2}}-3x+\dfrac{4}{x}+2$
(e)$h\left( x \right)={{x}^{4}}-{{x}^{\dfrac{3}{2}}}+x-1$
(f)$f\left( x \right)=2+\dfrac{3}{x}+4x$
Answer
596.7k+ views
Hint: First we are going to look at the definition of polynomials and with the help of that we are going to determine which satisfies the condition to be a polynomial or not.
Complete step-by-step answer:
Let’s first write the definition of polynomial and then we will use that to check all the given options:
Polynomial: In mathematics, a polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
Hence from this definition of polynomial it is very clear that the power of a variable must be non-negative integer.
Let check all the options one by one,
For (a): $f\left( x \right)=4{{x}^{3}}-{{x}^{2}}-3x+7$
Here all the power of x is non-negative integer.
Hence it is a polynomial function.
For (b): $g\left( x \right)=2{{x}^{3}}-3{{x}^{2}}+\sqrt{x}-1$
In $\sqrt{x}$ the power of x is $\dfrac{1}{2}$ which is not an integer.
Hence it is not a polynomial function.
For (c): $p\left( x \right)=\dfrac{2{{x}^{2}}}{3}-\dfrac{7x}{4}+9$
Here all the power of x is non-negative integer.
Hence it is a polynomial function.
For (d): $q\left( x \right)=2{{x}^{2}}-3x+\dfrac{4}{x}+2$
In $\dfrac{4}{x}$ the power of x is -1 which is a negative integer.
Hence it is not a polynomial function.
For (e): $h\left( x \right)={{x}^{4}}-{{x}^{\dfrac{3}{2}}}+x-1$
In ${{x}^{\dfrac{3}{2}}}$ the power of x is $\dfrac{3}{2}$ which is not an integer.
Hence it is not a polynomial function.
For (f): $f\left( x \right)=2+\dfrac{3}{x}+4x$
In $\dfrac{3}{x}$ the power of x is -1 which is a negative integer.
Hence it is not a polynomial function.
Note: Here we have used the definition of polynomials to solve this question. The definition should be clear to the student to solve this question correctly either it will lead to mistakes.
Complete step-by-step answer:
Let’s first write the definition of polynomial and then we will use that to check all the given options:
Polynomial: In mathematics, a polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
Hence from this definition of polynomial it is very clear that the power of a variable must be non-negative integer.
Let check all the options one by one,
For (a): $f\left( x \right)=4{{x}^{3}}-{{x}^{2}}-3x+7$
Here all the power of x is non-negative integer.
Hence it is a polynomial function.
For (b): $g\left( x \right)=2{{x}^{3}}-3{{x}^{2}}+\sqrt{x}-1$
In $\sqrt{x}$ the power of x is $\dfrac{1}{2}$ which is not an integer.
Hence it is not a polynomial function.
For (c): $p\left( x \right)=\dfrac{2{{x}^{2}}}{3}-\dfrac{7x}{4}+9$
Here all the power of x is non-negative integer.
Hence it is a polynomial function.
For (d): $q\left( x \right)=2{{x}^{2}}-3x+\dfrac{4}{x}+2$
In $\dfrac{4}{x}$ the power of x is -1 which is a negative integer.
Hence it is not a polynomial function.
For (e): $h\left( x \right)={{x}^{4}}-{{x}^{\dfrac{3}{2}}}+x-1$
In ${{x}^{\dfrac{3}{2}}}$ the power of x is $\dfrac{3}{2}$ which is not an integer.
Hence it is not a polynomial function.
For (f): $f\left( x \right)=2+\dfrac{3}{x}+4x$
In $\dfrac{3}{x}$ the power of x is -1 which is a negative integer.
Hence it is not a polynomial function.
Note: Here we have used the definition of polynomials to solve this question. The definition should be clear to the student to solve this question correctly either it will lead to mistakes.
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