Let
`P(x)=|(7,6,x-10),(2,x-10,5),(x-10,3,4)|`
sum of zeros of P(x) is

Options

A. 30
B. 28
C. 27
D. 25

Answer

To find the sum of the zeros of the polynomial \( P(x) = \begin{vmatrix} 7 & 6 & x-10 \\ 2 & x-10 & 5 \\ x-10 & 3 & 4 \end{vmatrix} \), we will first compute the determinant and then analyze the resulting polynomial.

### Step 1: Compute the Determinant

We can calculate the determinant using the formula for a \( 3 \times 3 \) matrix:

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

Where the matrix is:

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

For our matrix:

\[ A = \begin{pmatrix} 7 & 6 & x-10 \\ 2 & x-10 & 5 \\ x-10 & 3 & 4 \end{pmatrix} \]

We identify:

- \( a = 7 \), \( b = 6 \), \( c = x-10 \)
- \( d = 2 \), \( e = x-10 \), \( f = 5 \)
- \( g = x-10 \), \( h = 3 \), \( i = 4 \)

Now we can compute the determinant:

\[ P(x) = 7 \begin{vmatrix} x-10 & 5 \\ 3 & 4 \end{vmatrix} - 6 \begin{vmatrix} 2 & 5 \\ x-10 & 4 \end{vmatrix} + (x-10) \begin{vmatrix} 2 & x-10 \\ x-10 & 3 \end{vmatrix} \]

### Step 2: Calculate Each Minor Determinant

1. **First Minor:**

\[ \begin{vmatrix} x-10 & 5 \\ 3 & 4 \end{vmatrix} = (x-10) \cdot 4 - 5 \cdot 3 = 4(x-10) - 15 = 4x - 40 - 15 = 4x - 55 \]

2. **Second Minor:**

\[ \begin{vmatrix} 2 & 5 \\ x-10 & 4 \end{vmatrix} = 2 \cdot 4 - 5 \cdot (x-10) = 8 - 5(x-10) = 8 - 5x + 50 = -5x + 58 \]

3. **Third Minor:**

\[ \begin{vmatrix} 2 & x-10 \\ x-10 & 3 \end{vmatrix} = 2 \cdot 3 - (x-10)(x-10) = 6 - (x-10)^2 = 6 - (x^2 - 20x + 100) = -x^2 + 20x - 94 \]

### Step 3: Substitute Back into the Determinant Equation

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

\[ P(x) = 7(4x - 55) - 6(-5x + 58) + (x-10)(-x^2 + 20x - 94) \]

Calculating each term:

1. \( 7(4x - 55) = 28x - 385 \)
2. \( -6(-5x + 58) = 30x - 348 \)
3. Expanding \( (x-10)(-x^2 + 20x - 94) = -x^3 + 20x^2 - 94x + 10x^2 - 940 = -x^3 + 30x^2 - 94x - 940 \)

### Step 4: Combine All Terms

Now combine all terms:

\[ P(x) = (-x^3) + (28x + 30x - 94x + 30x^2) + (-385 - 348 - 940) \]

This simplifies to:

\[ P(x) = -x^3 + 30x^2 - 236x - 1673 \]

### Step 5: Find the Sum of the Zeros

The sum of the zeros of a polynomial \( ax^3 + bx^2 + cx + d \) is given by:

\[ -\frac{b}{a} \]

Here, \( a = -1 \) and \( b = 30 \):

\[ \text{Sum of zeros} = -\frac{30}{-1} = 30 \]

### Final Answer

The sum of the zeros of \( P(x) \) is \( \boxed{30} \).