`|(1+x,2,3,4),(1,2+x,3,4),(1,2,3+x,4),(1,2,3,4+x)|` =

Options

A. `x(x+10)`
B. `x^2(x+10)`
C. `x^3(x+10)`
D. none

Answer

To solve the determinant \( D = \begin{vmatrix} 1+x & 2 & 3 & 4 \\ 1 & 2+x & 3 & 4 \\ 1 & 2 & 3+x & 4 \\ 1 & 2 & 3 & 4+x \end{vmatrix} \), we will perform a series of row and column operations to simplify the determinant.

### Step 1: Row Operations
We will perform the operation \( R_1 - R_2 \) and \( R_1 - R_3 \) and \( R_1 - R_4 \) to simplify the first row.

\[ R_1 \rightarrow R_1 - R_2 \]
This gives:
\[ \begin{vmatrix} x & 0 & 0 & 0 \\ 1 & 2+x & 3 & 4 \\ 1 & 2 & 3+x & 4 \\ 1 & 2 & 3 & 4+x \end{vmatrix} \]

### Step 2: Further Row Operations
Next, we will perform \( R_2 - R_3 \) and \( R_2 - R_4 \).

\[ R_2 \rightarrow R_2 - R_3 \]
This gives:
\[ \begin{vmatrix} x & 0 & 0 & 0 \\ 0 & x & 0 & 0 \\ 1 & 2 & 3+x & 4 \\ 1 & 2 & 3 & 4+x \end{vmatrix} \]

### Step 3: Column Operations
Now, we will perform column operations to simplify the determinant further. We will perform \( C_2 \rightarrow C_2 - C_1 \).

\[ C_2 \rightarrow C_2 - C_1 \]
This gives:
\[ \begin{vmatrix} x & 0 & 0 & 0 \\ 0 & x & 0 & 0 \\ 1 & 1 & 3+x & 4 \\ 1 & 1 & 3 & 4+x \end{vmatrix} \]

### Step 4: Expanding the Determinant
Now we can expand the determinant along the first row:
\[ D = x \cdot \begin{vmatrix} x & 0 & 0 \\ 1 & 3+x & 4 \\ 1 & 3 & 4+x \end{vmatrix} \]

### Step 5: Calculate the 3x3 Determinant
Now we need to calculate the 3x3 determinant:
\[ \begin{vmatrix} x & 0 & 0 \\ 1 & 3+x & 4 \\ 1 & 3 & 4+x \end{vmatrix} \]
This determinant can be simplified to:
\[ x \cdot \begin{vmatrix} 3+x & 4 \\ 3 & 4+x \end{vmatrix} \]

### Step 6: Calculate the 2x2 Determinant
Calculating the 2x2 determinant:
\[ \begin{vmatrix} 3+x & 4 \\ 3 & 4+x \end{vmatrix} = (3+x)(4+x) - (3)(4) = 12 + 7x + x^2 - 12 = x^2 + 7x \]

### Step 7: Final Expression
Thus, we have:
\[ D = x \cdot x \cdot (x^2 + 7x) = x^2(x^2 + 7x) = x^4 + 7x^3 \]

### Final Answer
The value of the determinant is:
\[ D = x^4 + 7x^3 \]