If the slope of the line through the origin which is tangent to the curve `y=x^(3)+x+16` is `m`, then the value of `m-12` is

Explanation

To solve the problem, we need to find the slope \( m \) of the line through the origin that is tangent to the curve given by the equation \( y = x^3 + x + 16 \). We will then compute \( m - 12 \).

### Step-by-Step Solution:

1. **Differentiate the Curve**:
The first step is to find the derivative of the curve to determine the slope of the tangent line at any point \( x \).
\[ \frac{dy}{dx} = 3x^2 + 1 \]

2. **Set the Point of Tangency**:
Let the point of tangency be \( (x_1, y_1) \). The coordinates of this point can be expressed as:
\[ y_1 = x_1^3 + x_1 + 16 \]

3. **Slope of the Tangent Line**:
The slope \( m \) of the tangent line at the point \( (x_1, y_1) \) is given by:
\[ m = \frac{y_1 - 0}{x_1 - 0} = \frac{y_1}{x_1} \]
Substituting \( y_1 \):
\[ m = \frac{x_1^3 + x_1 + 16}{x_1} \]

4. **Simplify the Expression for \( m \)**:
We can simplify the expression for \( m \):
\[ m = x_1^2 + 1 + \frac{16}{x_1} \]

5. **Equate the Two Expressions for \( m \)**:
From the derivative, we also have:
\[ m = 3x_1^2 + 1 \]
Now, we can set the two expressions for \( m \) equal to each other:
\[ x_1^2 + 1 + \frac{16}{x_1} = 3x_1^2 + 1 \]

6. **Rearranging the Equation**:
Subtract \( 1 \) from both sides:
\[ x_1^2 + \frac{16}{x_1} = 3x_1^2 \]
Rearranging gives:
\[ 16 = 3x_1^2 - x_1^2 \]
Simplifying further:
\[ 16 = 2x_1^2 \]
Thus:
\[ x_1^2 = 8 \quad \Rightarrow \quad x_1 = 2 \quad (\text{since } x_1 \text{ must be positive}) \]

7. **Finding the Value of \( m \)**:
Now we substitute \( x_1 = 2 \) back into the expression for \( m \):
\[ m = 3(2^2) + 1 = 3(4) + 1 = 12 + 1 = 13 \]

8. **Calculate \( m - 12 \)**:
Finally, we compute \( m - 12 \):
\[ m - 12 = 13 - 12 = 1 \]

### Final Answer:
The value of \( m - 12 \) is \( \boxed{1} \).