Find all integers $ k$ such that for infinitely many integers $ n \ge 3$ the polynomial
\[ P(x) =x^{n+ 1}+ kx^n - 870x^2 + 1945x + 1995\]
can be reduced into two polynomials with integer coefficients.
1. **Initial Considerations**:
   - We need to find all integers \( k \) such that for infinitely many integers \( n \ge 3 \), the polynomial 
     \[
     P(x) = x^{n+1} + kx^n - 870x^2 + 1945x + 1995
     \]
     can be factored into two polynomials with integer coefficients.
   - If \( k \) were divisible by 5, Eisenstein's criterion would imply that \( P(x) \) is irreducible in \( \mathbb{Z}[x] \). Therefore, \( k \) is not divisible by 5.

2. **Factorization Form**:
   - If \( P(x) \) is reducible, it can be written as:
     \[
     P(x) = (a_k x^k + \cdots + a_1 x + a_0)(b_j x^j + \cdots + b_1 x + b_0)
     \]
     where \( k + j = n + 1 \).

3. **Constant Term Analysis**:
   - The constant term of \( P(x) \) is \( a_0 b_0 = 1995 \).
   - Since 1995 is not divisible by 25, only one of \( a_0 \) or \( b_0 \) can be divisible by 5. Assume \( a_0 = 5a_0' \).

4. **Extremal Principle**:
   - Let \( i \) be the smallest integer such that \( a_i \) is not divisible by 5.
   - Consider the coefficient of \( x^i \) in the expansion. It is \( a_i b_0 + a_{i-1} b_1 + \cdots \).
   - Since all \( a_g \) for \( g < i \) are divisible by 5, the coefficient is congruent to \( a_i b_0 \pmod{5} \).
   - Since neither \( a_i \) nor \( b_0 \) are divisible by 5, the coefficient of \( x^i \) is not divisible by 5.

5. **Coefficient Analysis**:
   - The only coefficients in \( P(x) \) not divisible by 5 are \( kx^n \) and \( x^{n+1} \).
   - Therefore, \( i = n \) or \( n+1 \). Clearly, \( i \neq n+1 \) because then \( k = n+1 \) implies \( j = 0 \), making \( b_j x^j + \cdots + b_1 x + b_0 \) a constant.
   - Thus, \( i = n \) and by similar reasoning, \( k = n \) implies \( j = 1 \).

6. **Rational Root Theorem**:
   - Since \( b_1 x + b_0 \) must divide \( P(x) \), \( P(x) \) must have a rational root.
   - By the Rational Root Theorem, the possible rational roots of \( P(x) \) are the divisors of 1995.
   - Suppose \( d \) is a root of \( P(x) \) for infinitely many \( n \). If \( |d| > 1 \), for large \( n \), \( |d^{n+1} + kd^n| > |870d^2 - 1945d - 1995| \), implying \( P(d) \neq 0 \).
   - Therefore, \( d = \pm 1 \) for infinitely many \( n \).

7. **Solving for \( k \)**:
   - If \( d = 1 \):
     \[
     1 + k - 870 + 1945 + 1995 = 0 \implies k = -3071
     \]
   - If \( d = -1 \):
     \[
     (-1)^{n+1} + k(-1)^n - 870(-1)^2 + 1945(-1) + 1995 = 0
     \]
     - For \( n \) even:
       \[
       -1 + k - 870 - 1945 + 1995 = 0 \implies k = -821
       \]
     - For \( n \) odd:
       \[
       1 - k - 870 + 1945 + 1995 = 0 \implies k = 821
       \]

Thus, the only possible values for \( k \) are \( -3071, -821, \) and \( 821 \).

The final answer is \( \boxed{ k = -3071, -821, 821 } \)