Let
`P(x)=|(x+1,2,3),(1,x+2,3),(1,2,x+3)|` the product of zeros of P(x) is

Options

A. 0
B. 6
C. `-6`
D. 12

Answer

To find the product of the zeros of the polynomial \( P(x) = |(x+1, 2, 3), (1, x+2, 3), (1, 2, x+3)| \), we will first compute the determinant and then identify the product of its roots.

### Step 1: Calculate the Determinant

The determinant \( P(x) \) can be calculated using the formula for a 3x3 determinant:

\[ P(x) = \begin{vmatrix} x+1 & 2 & 3 \\ 1 & x+2 & 3 \\ 1 & 2 & x+3 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where the matrix is:

\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

### Step 2: Apply the Determinant Formula

Here, we have:

- \( a = x+1, b = 2, c = 3 \)
- \( d = 1, e = x+2, f = 3 \)
- \( g = 1, h = 2, i = x+3 \)

Now substituting into the determinant formula:

\[ P(x) = (x+1)((x+2)(x+3) - 3 \cdot 2) - 2(1 \cdot (x+3) - 3 \cdot 1) + 3(1 \cdot 2 - 1 \cdot (x+2)) \]

### Step 3: Simplify Each Term

1. **Calculate \( (x+2)(x+3) - 6 \)**:
\[ (x+2)(x+3) = x^2 + 5x + 6 \implies (x^2 + 5x + 6 - 6) = x^2 + 5x \]

2. **Calculate \( 2(1 \cdot (x+3) - 3) \)**:
\[ 1 \cdot (x+3) - 3 = x + 3 - 3 = x \implies -2x \]

3. **Calculate \( 3(2 - (x+2)) \)**:
\[ 2 - (x+2) = 2 - x - 2 = -x \implies 3(-x) = -3x \]

### Step 4: Combine the Results

Now combine all the parts together:

\[ P(x) = (x+1)(x^2 + 5x) - 2x - 3x \]
\[ = (x+1)(x^2 + 5x) - 5x \]

Expanding \( (x+1)(x^2 + 5x) \):

\[ = x^3 + 5x^2 + x^2 + 5x = x^3 + 6x^2 + 5x \]

Now, substituting back into \( P(x) \):

\[ P(x) = x^3 + 6x^2 + 5x - 5x = x^3 + 6x^2 \]

### Step 5: Identify the Coefficients

The polynomial can be rewritten as:

\[ P(x) = x^3 + 6x^2 + 0x + 0 \]

### Step 6: Find the Product of the Zeros

For a cubic polynomial \( ax^3 + bx^2 + cx + d \), the product of the roots is given by:

\[ \text{Product of roots} = -\frac{d}{a} \]

Here, \( a = 1 \) and \( d = 0 \):

\[ \text{Product of roots} = -\frac{0}{1} = 0 \]

### Final Answer

The product of the zeros of \( P(x) \) is **0**.

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