m is the smallest positive integer such that for any integer `n le m`, the quantity `n^(3) - 7n^(2) + 11 n - 5` is positive. What is the value of m ?

Options

A. 4
B. 5
C. 8
D. None of these

Answer

To find the smallest positive integer \( m \) such that for any integer \( n \leq m \), the expression \( n^3 - 7n^2 + 11n - 5 \) is positive, we will analyze the polynomial step by step.

### Step 1: Identify the Polynomial
The polynomial we are dealing with is:
\[ P(n) = n^3 - 7n^2 + 11n - 5 \]

### Step 2: Find the Roots of the Polynomial
To determine when this polynomial is positive, we first need to find its roots. We can use the Rational Root Theorem to test possible rational roots, which are the factors of the constant term (-5) divided by the leading coefficient (1). The possible rational roots are \( \pm 1, \pm 5 \).

Let's test \( n = 1 \):
\[ P(1) = 1^3 - 7 \cdot 1^2 + 11 \cdot 1 - 5 = 1 - 7 + 11 - 5 = 0 \]
Since \( P(1) = 0 \), \( n - 1 \) is a factor of the polynomial.

### Step 3: Perform Polynomial Long Division
Now we will divide \( P(n) \) by \( n - 1 \):
\[ P(n) = (n - 1)(n^2 - 6n + 5) \]

### Step 4: Factor the Quadratic
Next, we need to factor the quadratic \( n^2 - 6n + 5 \):
\[ n^2 - 6n + 5 = (n - 1)(n - 5) \]
Thus, we can rewrite the polynomial as:
\[ P(n) = (n - 1)^2(n - 5) \]

### Step 5: Determine the Sign of the Polynomial
The polynomial \( P(n) = (n - 1)^2(n - 5) \) is positive when:
1. \( (n - 1)^2 \) is always non-negative (it is zero at \( n = 1 \)).
2. \( (n - 5) > 0 \) when \( n > 5 \).

### Step 6: Find the Smallest \( m \)
To ensure \( P(n) > 0 \) for all integers \( n \leq m \), we need \( n \) to be greater than 5. Therefore, the smallest integer \( m \) such that \( n \leq m \) and \( P(n) > 0 \) for all \( n \leq m \) is:
\[ m = 6 \]

### Conclusion
Thus, the value of \( m \) is:
\[ \boxed{6} \]