Determine if the given set is a subspace of set of prime numbers P 3. Justify your answer. The set of all polynomials of the form p (t) equalsat cubed , where a is in set of real numbers R.

It is not a subspace.

Step-by-step explanation:

So, the polynomial degree of at most 3 is given below as;

V = { p (z) = b0 + b1z + b2z^ + ... + bnz^n. |n is less than or equal to 3 and b0, b1, b2, ... are integers.

To determine whether a subset is a subspace of Pn, we have to check for the properties below;

(1). Zero vector property : that is, when polynomial, p (z) = 0 and 0 is an integer.

(2). Addition property = here, we have; p (z) + h (z) = (b0 + b1z + b2z^2 + ... + bnz^n) + (c0 + c1z + c^2z^2 + ... + cnz^n). That is the sum of integers.

(3). Scaler multiplication property: the coefficient here may not be real numbers therefore, the condition is not followed here.

Therefore, it is not a subspace of Pn.