If `f(x) = ax^4+bx^3+cx^2+dx +e`
`=|(x^3+3x,x-1,x+3),(x+1,-2x,x-4),(x-3,x+4,3x)|` then e=

Options

A. 1
B. `-1`
C. 2
D. 0

Answer

To find the value of \( e \) in the polynomial \( f(x) = ax^4 + bx^3 + cx^2 + dx + e \) given that it is equal to the determinant

\[ \begin{vmatrix} x^3 + 3x & x - 1 & x + 3 \\ x + 1 & -2x & x - 4 \\ x - 3 & x + 4 & 3x \end{vmatrix} \]

we can follow these steps:

### Step 1: Substitute \( x = 0 \)

We will evaluate \( f(0) \) to find \( e \).

\[ f(0) = e \]

Now, we substitute \( x = 0 \) into the determinant:

\[ \begin{vmatrix} 0^3 + 3(0) & 0 - 1 & 0 + 3 \\ 0 + 1 & -2(0) & 0 - 4 \\ 0 - 3 & 0 + 4 & 3(0) \end{vmatrix} = \begin{vmatrix} 0 & -1 & 3 \\ 1 & 0 & -4 \\ -3 & 4 & 0 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Now we need to calculate the determinant:

\[ D = \begin{vmatrix} 0 & -1 & 3 \\ 1 & 0 & -4 \\ -3 & 4 & 0 \end{vmatrix} \]

We can expand this determinant along the first row:

\[ D = 0 \cdot \begin{vmatrix} 0 & -4 \\ 4 & 0 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -4 \\ -3 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 0 \\ -3 & 4 \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants

Now we calculate the 2x2 determinants:

1. For the second term:
\[ \begin{vmatrix} 1 & -4 \\ -3 & 0 \end{vmatrix} = (1)(0) - (-4)(-3) = 0 - 12 = -12 \]

2. For the third term:
\[ \begin{vmatrix} 1 & 0 \\ -3 & 4 \end{vmatrix} = (1)(4) - (0)(-3) = 4 - 0 = 4 \]

### Step 4: Substitute Back into the Determinant

Now substituting back into the determinant calculation:

\[ D = 0 - (-1)(-12) + 3(4) = 0 - 12 + 12 = 0 \]

### Step 5: Conclusion

Thus, we find that:

\[ e = D = 0 \]

### Final Answer

The value of \( e \) is \( \boxed{0} \).

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