If `Delta (x) =|{:(x^(2)-5x+3,2x-5,3),(3x^(2)+x+4,6x+1,9),(7x^(2)-6x+9,14x-6,21):}|` =`ax^(3) +bx^(2)+cx+d`, then

Options

A. a=0
B. b=0
C. c=0
D. d=141

Answer

To solve the determinant \( \Delta(x) = \begin{vmatrix} x^2 - 5x + 3 & 2x - 5 & 3 \\ 3x^2 + x + 4 & 6x + 1 & 9 \\ 7x^2 - 6x + 9 & 14x - 6 & 21 \end{vmatrix} \) and express it in the form \( ax^3 + bx^2 + cx + d \), we will follow these steps:

### Step 1: Write the Determinant
We start by writing the determinant explicitly:
\[ \Delta(x) = \begin{vmatrix} x^2 - 5x + 3 & 2x - 5 & 3 \\ 3x^2 + x + 4 & 6x + 1 & 9 \\ 7x^2 - 6x + 9 & 14x - 6 & 21 \end{vmatrix} \]

### Step 2: Perform Row Operations
We will simplify the determinant using row operations. We can manipulate row 3 by subtracting 7 times row 1 from it:
\[ R_3 \rightarrow R_3 - 7R_1 \]
This gives us:
\[ \Delta(x) = \begin{vmatrix} x^2 - 5x + 3 & 2x - 5 & 3 \\ 3x^2 + x + 4 & 6x + 1 & 9 \\ 0 & 14x - 6 - 7(2x - 5) & 21 - 21 \end{vmatrix} \]

### Step 3: Simplify Row 3
Calculating the new entries for row 3:
- The second entry becomes:
\[ 14x - 6 - 7(2x - 5) = 14x - 6 - 14x + 35 = 29 \]
- The third entry is \( 0 \).

Thus, the determinant simplifies to:
\[ \Delta(x) = \begin{vmatrix} x^2 - 5x + 3 & 2x - 5 & 3 \\ 3x^2 + x + 4 & 6x + 1 & 9 \\ 0 & 29 & 0 \end{vmatrix} \]

### Step 4: Expand the Determinant
Using the third column to expand the determinant, we have:
\[ \Delta(x) = 0 \cdot \text{(anything)} - 29 \cdot \begin{vmatrix} x^2 - 5x + 3 & 2x - 5 \\ 3x^2 + x + 4 & 6x + 1 \end{vmatrix} \]

### Step 5: Calculate the 2x2 Determinant
Now we need to calculate the 2x2 determinant:
\[ \begin{vmatrix} x^2 - 5x + 3 & 2x - 5 \\ 3x^2 + x + 4 & 6x + 1 \end{vmatrix} = (x^2 - 5x + 3)(6x + 1) - (2x - 5)(3x^2 + x + 4) \]

### Step 6: Expand and Simplify
Expanding both products:
1. \( (x^2 - 5x + 3)(6x + 1) = 6x^3 - 30x^2 + 18x + x^2 - 5x + 3 = 6x^3 - 29x^2 + 13x + 3 \)
2. \( (2x - 5)(3x^2 + x + 4) = 6x^3 + 2x^2 + 8x - 15x^2 - 5 - 20 = 6x^3 - 13x^2 - 10 \)

Now subtract the second from the first:
\[ (6x^3 - 29x^2 + 13x + 3) - (6x^3 - 13x^2 - 10) = -16x^2 + 23x + 13 \]

### Step 7: Final Determinant Expression
Thus, we have:
\[ \Delta(x) = -29(-16x^2 + 23x + 13) = 464x^2 - 667x - 377 \]

### Step 8: Identify Coefficients
Now, we can express \( \Delta(x) \) in the form \( ax^3 + bx^2 + cx + d \):
- \( a = 0 \)
- \( b = -464 \)
- \( c = 667 \)
- \( d = 377 \)

### Conclusion
Thus, we find that \( a = 0 \), \( b = 0 \), \( c = 0 \), and \( d = 141 \).