Question

which of the following is a polynomial?
a. \\(\frac{1}{x} + 2\\)
b. \\(x^2 - 1\\)
c. \\(\frac{(x^6 - 2)}{(x^{-4} + 3)}\\)
d. \\(x^4 + x^{-4} + 16\\)

Explanation:

Step1: Recall Polynomial Definition

A polynomial is an expression of the form \(a_nx^n + a_{n - 1}x^{n - 1}+\dots+a_1x + a_0\), where \(n\) is a non - negative integer and \(a_i\) are constants. Exponents must be non - negative integers, and there are no variables in the denominator.

Step2: Analyze Option A

For option A, \(\frac{1}{x}+2=x^{- 1}+2\). The exponent of \(x\) is \(- 1\), which is a negative integer. So, it is not a polynomial.

Step3: Analyze Option B

For option B, \(x^{2}-1\) can be written as \(1\times x^{2}+0\times x+( - 1)\). Here, the exponents of \(x\) are \(2\) (a non - negative integer) and \(0\) (implicit for the constant term). So, it is a polynomial.

Step4: Analyze Option C

First, simplify \(\frac{x^{6}-2}{x^{-4}+3}\). Multiply the numerator and denominator by \(x^{4}\) to get \(\frac{(x^{6}-2)x^{4}}{1 + 3x^{4}}=\frac{x^{10}-2x^{4}}{3x^{4}+1}\). There is a variable in the denominator, so it is not a polynomial.

Step5: Analyze Option D

For option D, \(x^{4}+x^{-4}+16=x^{4}+\frac{1}{x^{4}}+16\). There is a variable in the denominator (since \(x^{-4}=\frac{1}{x^{4}}\)), so it is not a polynomial.

Answer:

B. \(x^{2}-1\)