Question

$\sqrt{2}$ is a polynomial of degree _____. select the correct response: 0, 2, $\sqrt{}$, 1/2, 1

Explanation:

Step1: Recall Polynomial Degree

A polynomial in one variable \( x \) is an expression of the form \( a_nx^n + a_{n - 1}x^{n - 1}+\dots+a_1x + a_0 \), where \( a_i \) are constants and \( n \) is a non - negative integer. The degree of a non - zero constant polynomial (a polynomial with no variable term, just a constant) is 0.

Step2: Analyze \( \sqrt{2} \) as a Polynomial

The number \( \sqrt{2} \) is a constant. We can think of it as a polynomial in \( x \) (for any variable \( x \)): \( \sqrt{2}= \sqrt{2}x^0 \) (since \( x^0 = 1 \) for \( x
eq0 \)). Comparing with the general form of a polynomial \( a_nx^n+\dots+a_0 \), here \( n = 0 \), \( a_0=\sqrt{2} \) and all other coefficients \( a_i = 0 \) for \( i>0 \). So the degree of the polynomial \( \sqrt{2} \) is 0.

Answer:

0 (corresponding to the option with the text "0")