If the slope of line through the origin which is tangent to the curve `y=x^3+x+16` is `m ,` then the value of `m-4` is____.

Answer

To solve the problem, we need to find the slope of the tangent line to the curve \( y = x^3 + x + 16 \) that passes through the origin. We will denote the point of tangency on the curve as \( (a, b) \).

### Step-by-Step Solution:

1. **Identify the curve and point of tangency**:
The curve is given by:
\[ y = x^3 + x + 16 \]
At the point \( (a, b) \), we have:
\[ b = a^3 + a + 16 \]

2. **Find the derivative of the curve**:
To find the slope of the tangent line at any point \( x \), we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3x^2 + 1 \]
Therefore, the slope of the tangent line at the point \( (a, b) \) is:
\[ m = 3a^2 + 1 \]

3. **Set up the slope condition**:
Since the tangent line passes through the origin \( (0, 0) \), the slope of the line from the origin to the point \( (a, b) \) is given by:
\[ \text{slope} = \frac{b - 0}{a - 0} = \frac{b}{a} \]
Thus, we have:
\[ \frac{b}{a} = m \]

4. **Substitute for \( b \)**:
From the previous steps, we know that:
\[ b = a^3 + a + 16 \]
Substituting this into the slope condition gives:
\[ \frac{a^3 + a + 16}{a} = 3a^2 + 1 \]

5. **Multiply through by \( a \)** (assuming \( a \neq 0 \)):
\[ a^3 + a + 16 = a(3a^2 + 1) \]
Simplifying the right side:
\[ a^3 + a + 16 = 3a^3 + a \]

6. **Rearranging the equation**:
Move all terms to one side:
\[ a^3 + a + 16 - 3a^3 - a = 0 \]
This simplifies to:
\[ -2a^3 + 16 = 0 \]
Thus:
\[ 2a^3 = 16 \quad \Rightarrow \quad a^3 = 8 \quad \Rightarrow \quad a = 2 \]

7. **Calculate the slope \( m \)**:
Now, substituting \( a = 2 \) back into the slope formula:
\[ m = 3(2^2) + 1 = 3(4) + 1 = 12 + 1 = 13 \]

8. **Find \( m - 4 \)**:
Finally, we need to calculate:
\[ m - 4 = 13 - 4 = 9 \]

### Final Answer:
The value of \( m - 4 \) is \( \boxed{9} \).