If `(n+1)! =12xx(n-1)!`, find the value of n.

Explanation

To solve the equation \((n+1)! = 12 \times (n-1)!\), we can follow these steps:

### Step 1: Rewrite the factorials
We know that:
\[ (n+1)! = (n+1) \times n \times (n-1)! \]
So we can rewrite the equation as:
\[ (n+1) \times n \times (n-1)! = 12 \times (n-1)! \]

### Step 2: Cancel out \((n-1)!\)
Since \((n-1)!\) is common on both sides of the equation and is not equal to zero for \(n \geq 2\), we can safely divide both sides by \((n-1)!\):
\[ (n+1) \times n = 12 \]

### Step 3: Expand the left side
Now, we can expand the left side:
\[ n^2 + n = 12 \]

### Step 4: Rearrange the equation
Rearranging the equation gives us a standard quadratic form:
\[ n^2 + n - 12 = 0 \]

### Step 5: Factor the quadratic equation
Next, we need to factor the quadratic equation. We are looking for two numbers that multiply to \(-12\) and add up to \(1\). The numbers \(4\) and \(-3\) fit this requirement:
\[ (n + 4)(n - 3) = 0 \]

### Step 6: Solve for \(n\)
Setting each factor to zero gives us:
\[ n + 4 = 0 \quad \Rightarrow \quad n = -4 \quad \text{(not valid since } n \text{ must be positive)} \]
\[ n - 3 = 0 \quad \Rightarrow \quad n = 3 \]

### Conclusion
Thus, the value of \(n\) is:
\[ \boxed{3} \]

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