Let $n$ be an integer with $n\geq 2.$ Over all real polynomials $p(x)$ of degree $n,$ what is the largest possible number of negative coefficients of $p(x)^2?$
1. **Claim**: The maximum number of negative coefficients of \( (P(x))^2 \) is \( 2n-2 \).

2. **Example for \( n = 2 \)**:
   - Consider a polynomial \( P(x) = a_2 x^2 + a_1 x + a_0 \).
   - Squaring \( P(x) \), we get:
     \[
     P(x)^2 = (a_2 x^2 + a_1 x + a_0)^2 = a_2^2 x^4 + 2a_1 a_2 x^3 + (2a_0 a_2 + a_1^2) x^2 + 2a_0 a_1 x + a_0^2
     \]
   - The coefficients are:
     \[
     \begin{aligned}
     &\text{Coefficient of } x^4: a_2^2 \geq 0 \\
     &\text{Coefficient of } x^3: 2a_1 a_2 \\
     &\text{Coefficient of } x^2: 2a_0 a_2 + a_1^2 \\
     &\text{Coefficient of } x: 2a_0 a_1 \geq 0 \\
     &\text{Constant term}: a_0^2 \geq 0
     \end{aligned}
     \]
   - To maximize the number of negative coefficients, we need \( 2a_1 a_2 < 0 \) and \( 2a_0 a_2 + a_1^2 < 0 \). This can be achieved by choosing appropriate signs for \( a_0, a_1, \) and \( a_2 \).

3. **Example for \( n = 3 \)**:
   - Consider a polynomial \( P(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0 \).
   - Squaring \( P(x) \), we get:
     \[
     P(x)^2 = (a_3 x^3 + a_2 x^2 + a_1 x + a_0)^2 = a_3^2 x^6 + 2a_3 a_2 x^5 + (a_2^2 + 2a_3 a_1) x^4 + (2a_0 a_3 + 2a_2 a_1) x^3 + (a_1^2 + 2a_0 a_2) x^2 + 2a_0 a_1 x + a_0^2
     \]
   - The coefficients are:
     \[
     \begin{aligned}
     &\text{Coefficient of } x^6: a_3^2 \geq 0 \\
     &\text{Coefficient of } x^5: 2a_3 a_2 \\
     &\text{Coefficient of } x^4: a_2^2 + 2a_3 a_1 \\
     &\text{Coefficient of } x^3: 2a_0 a_3 + 2a_2 a_1 \\
     &\text{Coefficient of } x^2: a_1^2 + 2a_0 a_2 \\
     &\text{Coefficient of } x: 2a_0 a_1 \\
     &\text{Constant term}: a_0^2 \geq 0
     \end{aligned}
     \]
   - To maximize the number of negative coefficients, we need \( 2a_3 a_2 < 0 \), \( a_2^2 + 2a_3 a_1 < 0 \), \( 2a_0 a_3 + 2a_2 a_1 < 0 \), and \( a_1^2 + 2a_0 a_2 < 0 \). This can be achieved by choosing appropriate signs for \( a_0, a_1, a_2, \) and \( a_3 \).

4. **General Case for \( n \)**:
   - Consider a polynomial \( P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \).
   - Squaring \( P(x) \), we get:
     \[
     P(x)^2 = \left( \sum_{i=0}^n a_i x^i \right)^2 = \sum_{k=0}^{2n} \left( \sum_{i+j=k} a_i a_j \right) x^k
     \]
   - The coefficient of \( x^k \) is \( \sum_{i+j=k} a_i a_j \).
   - To maximize the number of negative coefficients, we need to ensure that as many of these sums as possible are negative.

5. **Proof**:
   - We need to show that the maximum number of negative coefficients of \( P(x)^2 \) is \( 2n-2 \).
   - The coefficients of \( x^{2n} \) and \( x^{2n-1} \) are always non-negative because they involve squares and products of the highest degree terms.
   - For the remaining coefficients, we can choose the signs of \( a_0, a_1, \ldots, a_{n-1} \) such that \( 2n-2 \) of the sums \( \sum_{i+j=k} a_i a_j \) are negative.

6. **Conclusion**:
   - The maximum number of negative coefficients of \( P(x)^2 \) is \( 2n-2 \).

The final answer is \( \boxed{ 2n-2 } \).