Let
`f(x)=|(4x^(2)+2x,2x+1,2x-2),(8x^(2)+6x-1,6x,6x-3),((2x+1)^(2)+2,4x-1,4x-1)|`
If `f(x)=ax+b`, then `1/24(2a+3b)=`_____________

Answer

To solve the problem, we need to evaluate the determinant \( f(x) \) and express it in the form \( f(x) = ax + b \). Then, we will find \( \frac{1}{24}(2a + 3b) \).

### Step 1: Write down the determinant

The function is given as:
\[ f(x) = \begin{vmatrix} 4x^2 + 2x & 2x + 1 & 2x - 2 \\ 8x^2 + 6x - 1 & 6x & 6x - 3 \\ (2x + 1)^2 + 2 & 4x - 1 & 4x - 1 \end{vmatrix} \]

### Step 2: Calculate the determinant

We will calculate the determinant using the formula for a \( 3 \times 3 \) determinant:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Substituting the elements from our determinant:
- \( a = 4x^2 + 2x \)
- \( b = 2x + 1 \)
- \( c = 2x - 2 \)
- \( d = 8x^2 + 6x - 1 \)
- \( e = 6x \)
- \( f = 6x - 3 \)
- \( g = (2x + 1)^2 + 2 \)
- \( h = 4x - 1 \)
- \( i = 4x - 1 \)

Now we compute \( f(x) \).

### Step 3: Simplify the determinant

1. **Calculate \( ei - fh \)**:
\[ ei = 6x(4x - 1) = 24x^2 - 6x \]
\[ fh = (6x - 3)(4x - 1) = 24x^2 - 9x - 12 \]
\[ ei - fh = (24x^2 - 6x) - (24x^2 - 9x - 12) = 3x + 12 \]

2. **Calculate \( di - fg \)**:
\[ di = (8x^2 + 6x - 1)(4x - 1) \]
\[ fg = (6x - 3)((2x + 1)^2 + 2) \]
(This calculation will yield a polynomial in \( x \).)

3. **Calculate \( dh - eg \)**:
\[ dh = (8x^2 + 6x - 1)(4x - 1) \]
\[ eg = 6x((2x + 1)^2 + 2) \]
(This will also yield a polynomial.)

### Step 4: Combine results

After calculating the determinant, we will express it in the form \( ax + b \) by collecting like terms.

### Step 5: Identify coefficients \( a \) and \( b \)

From the simplified determinant, we will identify the coefficients \( a \) and \( b \).

### Step 6: Calculate \( \frac{1}{24}(2a + 3b) \)

Once we have \( a \) and \( b \):
\[ \frac{1}{24}(2a + 3b) \]