The greatest whole number, by which the expression `n^(4)+6n^(3)+11n^(2)+6n+24` is divisible for every natural number n, is

Options

A. 6
B. 24
C. 12
D. 48

Answer

To find the greatest whole number by which the expression \( n^4 + 6n^3 + 11n^2 + 6n + 24 \) is divisible for every natural number \( n \), we can follow these steps:

### Step 1: Evaluate the expression for \( n = 1 \)
Substituting \( n = 1 \) into the expression:

\[ 1^4 + 6 \cdot 1^3 + 11 \cdot 1^2 + 6 \cdot 1 + 24 \]

Calculating each term:

\[ 1 + 6 + 11 + 6 + 24 = 48 \]

So, when \( n = 1 \), the value of the expression is \( 48 \).

### Step 2: Evaluate the expression for \( n = 2 \)
Now, substituting \( n = 2 \):

\[ 2^4 + 6 \cdot 2^3 + 11 \cdot 2^2 + 6 \cdot 2 + 24 \]

Calculating each term:

\[ 16 + 6 \cdot 8 + 11 \cdot 4 + 12 + 24 \]
\[ 16 + 48 + 44 + 12 + 24 = 144 \]

So, when \( n = 2 \), the value of the expression is \( 144 \).

### Step 3: Find the greatest common divisor (GCD)
Now we have the values of the expression for \( n = 1 \) and \( n = 2 \):
- For \( n = 1 \), the value is \( 48 \).
- For \( n = 2 \), the value is \( 144 \).

Next, we need to find the GCD of \( 48 \) and \( 144 \).

The factors of \( 48 \) are \( 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 \).

The factors of \( 144 \) are \( 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 \).

The greatest common factor is \( 48 \).

### Conclusion
Thus, the greatest whole number by which the expression \( n^4 + 6n^3 + 11n^2 + 6n + 24 \) is divisible for every natural number \( n \) is:

\[ \boxed{48} \]