If the polynomial $ P\left(x\right)$ satisfies $ 2P\left(x\right) \equal{} P\left(x \plus{} 3\right) \plus{} P\left(x \minus{} 3\right)$ for every real number $ x$, degree of $ P\left(x\right)$ will be at most 

$\textbf{(A)}\ 0   \qquad\textbf{(B)}\ 1   \qquad\textbf{(C)}\ 2   \qquad\textbf{(D)}\ 3   \qquad\textbf{(E)}\ \text{None}$
1. Given the polynomial \( P(x) \) satisfies the functional equation:
   \[
   2P(x) = P(x + 3) + P(x - 3)
   \]
   for every real number \( x \).

2. Assume \( P(x) \) is a polynomial of degree \( n \). We will analyze the implications of the given functional equation on the degree of \( P(x) \).

3. Let \( P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \).

4. Substitute \( x \) with \( x + 3 \) and \( x - 3 \) in the polynomial:
   \[
   P(x + 3) = a_n (x + 3)^n + a_{n-1} (x + 3)^{n-1} + \cdots + a_1 (x + 3) + a_0
   \]
   \[
   P(x - 3) = a_n (x - 3)^n + a_{n-1} (x - 3)^{n-1} + \cdots + a_1 (x - 3) + a_0
   \]

5. Add \( P(x + 3) \) and \( P(x - 3) \):
   \[
   P(x + 3) + P(x - 3) = a_n \left( (x + 3)^n + (x - 3)^n \right) + a_{n-1} \left( (x + 3)^{n-1} + (x - 3)^{n-1} \right) + \cdots + 2a_0
   \]

6. Equate this to \( 2P(x) \):
   \[
   2P(x) = 2a_n x^n + 2a_{n-1} x^{n-1} + \cdots + 2a_0
   \]

7. For the equality \( 2P(x) = P(x + 3) + P(x - 3) \) to hold, the coefficients of the highest degree terms on both sides must match. This implies:
   \[
   2a_n x^n = a_n \left( (x + 3)^n + (x - 3)^n \right)
   \]

8. Consider the highest degree term \( x^n \) in \( (x + 3)^n + (x - 3)^n \):
   \[
   (x + 3)^n + (x - 3)^n = 2x^n + \text{lower degree terms}
   \]

9. Therefore, the highest degree term on the right-hand side is \( 2a_n x^n \), which matches the left-hand side. This implies that the lower degree terms must cancel out.

10. For the lower degree terms to cancel out, the polynomial must be of degree at most 1. This is because any higher degree polynomial would introduce non-canceling terms.

11. Thus, the polynomial \( P(x) \) must be of degree at most 1.

The final answer is \(\boxed{1}\)