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carlosego

For this case we have the following:

  • [tex]\frac{3}{x^{-1} }+ \frac{3}{2}[/tex] which can be rewritten as [tex]3x + \frac{3}{2}[/tex], is a polynomial.
  • [tex]x ^ 2 + x^{\frac{1}{2} }+ x + 6[/tex] is not a polynomial because it has a coefficient that is not integer.
  • [tex]x ^ 3-9[/tex] is a polynomial
  • [tex]\sqrt{16x^{4} }[/tex] that can be rewritten as [tex]4x ^ 2[/tex], is a polynomial

So:

[tex]x ^ 2 + x^{\frac{1}{2} }+ x + 6[/tex] is not a polynomial

Answer:

Option B

Answer:

The expression which is not a polynomial is:

            B)   [tex]x^2+x^{\dfrac{1}{2}}+x+6[/tex]

Step-by-step explanation:

We know that the general form of a polynomial expression is given by:

[tex]P(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+......+a_2x^2+a_1x+a_0[/tex]

i.e. P(x) is a polynomial of degree n  , where n belongs to natural numbers and [tex]a_i's[/tex] belong to the real numbers.

and [tex]a_n\neq 0[/tex]

A)

[tex]\dfrac{3}{x^{-1}}+\dfrac{3}{2}[/tex]

which could also be written by:

[tex]3x+\dfrac{3}{2}[/tex]

Hence, this expression is a polynomial expression.

Option: a is incorrect.

B)

[tex]x^2+x^{\dfrac{1}{2}}+x+6[/tex]

Since in the second term the power of x does not belong to the set of natural numbers.

Hence, option: b is not a polynomial expression.

Hence, the correct answer is option: b

C)

[tex]x^3-9[/tex]

Since, the expression matches the definition of the polynomial expression.

Hence, option: c is incorrect.

D)

[tex]\sqrt{16x^4}[/tex]

This could also be written as:

[tex]=4x^2[/tex]

Hence, it is a polynomial expression.

Hence, option: d is incorrect.

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