Question

p (x)= root xcube + 1 is not a polynomial. give reason​

Answer 1

Explanation:

$\sqrt{ {x}^{3} + 1} \: \: is \: not \: a \: polynomial \: because$

$p(x) = \sqrt{ {x}^{3} + 1} = \sqrt{ {x}^{3} } + 1 = {x}^{ \frac{3}{2} } + 1$

As per the definition of polynomial, the power of variable term is always a positive integer. But here the power of x is 1/2, which is clearly not an integer. So, p(x) is not a polynomial function.

Answer provided by GauthMath.