Question:
$\sqrt{2}$ is a polynomial of degree
(a) 2
(b) 0
(C) 1
(d) $1 / 2$
Solution:
(b) $\sqrt{2}=-\sqrt{2} x^{\circ}$. Hence, $\sqrt{2}$ is a polynomial of degree 0 , because exponent of $x$ is 0 .
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