`(ax+7)(bx-1)=12x^(2)+kx+(b-13)`
If the equation above is true for all values of x where a, b, and k are non-zero constants, what is the value of k?

Options

A. `40`
B. `25`
C. `17`
D. `8`

Answer

To solve the equation \((ax + 7)(bx - 1) = 12x^2 + kx + (b - 13)\), we will follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the left-hand side of the equation:

\[ (ax + 7)(bx - 1) = abx^2 - ax + 7bx - 7 \]

Combining like terms, we get:

\[ abx^2 + (7b - a)x - 7 \]

### Step 2: Set the expanded form equal to the right-hand side
Now we set the expanded left-hand side equal to the right-hand side of the equation:

\[ abx^2 + (7b - a)x - 7 = 12x^2 + kx + (b - 13) \]

### Step 3: Compare coefficients
Since the equation holds for all values of \(x\), we can compare the coefficients of \(x^2\), \(x\), and the constant terms on both sides.

1. Coefficient of \(x^2\):
\[ ab = 12 \quad \text{(1)} \]

2. Coefficient of \(x\):
\[ 7b - a = k \quad \text{(2)} \]

3. Constant term:
\[ -7 = b - 13 \quad \text{(3)} \]

### Step 4: Solve for \(b\) from equation (3)
From equation (3), we can solve for \(b\):

\[ b - 13 = -7 \implies b = 6 \]

### Step 5: Substitute \(b\) into equation (1) to find \(a\)
Now we substitute \(b = 6\) into equation (1):

\[ a(6) = 12 \implies a = \frac{12}{6} = 2 \]

### Step 6: Substitute \(a\) and \(b\) into equation (2) to find \(k\)
Now we substitute \(a = 2\) and \(b = 6\) into equation (2):

\[ k = 7(6) - 2 = 42 - 2 = 40 \]

### Conclusion
Thus, the value of \(k\) is:

\[ \boxed{40} \]