Let `ax^3+bx^2+cx+d=|{:(3x,x+1,x-1),(x-3,-2x,x+2),(x+3,x-4,5x):}|` then the value of d is

Options

A. 5
B. 0
C. -6
D. 4

Answer

To find the value of \( d \) in the polynomial \( ax^3 + bx^2 + cx + d = |{:(3x,x+1,x-1),(x-3,-2x,x+2),(x+3,x-4,5x):}| \), we can evaluate the determinant by substituting \( x = 0 \).

### Step-by-Step Solution:

1. **Substitute \( x = 0 \)**:
\[ d = |{:(3(0),0+1,0-1),(0-3,-2(0),0+2),(0+3,0-4,5(0)):}| \]
This simplifies to:
\[ d = |{:(0,1,-1),(-3,0,2),(3,-4,0):}| \]

2. **Set up the determinant**:
\[ d = \begin{vmatrix} 0 & 1 & -1 \\ -3 & 0 & 2 \\ 3 & -4 & 0 \end{vmatrix} \]

3. **Calculate the determinant**:
Using the formula for the determinant of a 3x3 matrix:
\[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where:
- \( a = 0, b = 1, c = -1 \)
- \( d = -3, e = 0, f = 2 \)
- \( g = 3, h = -4, i = 0 \)

Plugging in the values:
\[ d = 0 \cdot (0 \cdot 0 - 2 \cdot -4) - 1 \cdot (-3 \cdot 0 - 2 \cdot 3) + (-1) \cdot (-3 \cdot -4 - 0 \cdot 3) \]

This simplifies to:
\[ d = 0 - 1 \cdot (0 - 6) - 1 \cdot (12 - 0) \]
\[ d = 0 + 6 - 12 \]
\[ d = -6 \]

4. **Conclusion**:
The value of \( d \) is \( -6 \).