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Is [tex]\(\sqrt{2}\)[/tex] a polynomial of degree?

a) 2
b) 0
c) 1
d) [tex]\(\frac{1}{2}\)[/tex]


Sagot :

Let's analyze the given options to determine the degree of the polynomial for [tex]\(\sqrt{2}\)[/tex].

A polynomial is an expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of the variables. A polynomial can be written in the general form:
[tex]\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \][/tex]

The degree of a polynomial is the highest power of the variable [tex]\(x\)[/tex] that appears with a non-zero coefficient.

Let's look at [tex]\(\sqrt{2}\)[/tex]. In this case:

1. [tex]\(\sqrt{2}\)[/tex] is a constant because it does not involve any variables.
2. Constants are considered polynomials where the variable [tex]\(x\)[/tex] is raised to the power of 0 because they can be written as [tex]\(a_0 x^0\)[/tex], where [tex]\(a_0 = \sqrt{2}\)[/tex] and [tex]\(x^0 = 1\)[/tex].
3. The highest power of the variable in this expression is 0.

Therefore, [tex]\(\sqrt{2}\)[/tex] is a polynomial of degree [tex]\(0\)[/tex].

So, the correct answer is:
[tex]\[ \boxed{0} \][/tex]