Prove that `(x-2)(x-1)` is factor of `|(1,1,x),(beta+1,beta+1,beta+x),(3,x+1,x+2)|` and hence find the quotient.

Answer

To prove that \((x-2)(x-1)\) is a factor of the determinant

\[ D = \begin{vmatrix} 1 & 1 & x \\ \beta + 1 & \beta + 1 & \beta + x \\ 3 & x + 1 & x + 2 \end{vmatrix} \]

and to find the quotient, we will follow these steps:

### Step 1: Calculate the Determinant

We start by calculating the determinant \(D\):

\[ D = 1 \cdot \begin{vmatrix} \beta + 1 & \beta + x \\ x + 1 & x + 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} \beta + 1 & \beta + x \\ 3 & x + 2 \end{vmatrix} + x \cdot \begin{vmatrix} \beta + 1 & \beta + 1 \\ 3 & x + 1 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

Now we compute the 2x2 determinants:

1. For the first determinant:

\[ \begin{vmatrix} \beta + 1 & \beta + x \\ x + 1 & x + 2 \end{vmatrix} = (\beta + 1)(x + 2) - (\beta + x)(x + 1) = \beta x + 2\beta + x + 2 - (\beta x + \beta + x^2 + x) = 2\beta + 2 - \beta - x^2 = \beta + 2 - x^2 \]

2. For the second determinant:

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

3. For the third determinant:

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

### Step 3: Substitute Back into the Determinant

Substituting these values back into \(D\):

\[ D = 1 \cdot (\beta + 2 - x^2) - 1 \cdot (-2\beta - 2x + 2) + x \cdot (\beta + 1)(x - 2) \]

This simplifies to:

\[ D = \beta + 2 - x^2 + 2\beta + 2x - 2 + x(\beta + 1)(x - 2) \]

Combine like terms:

\[ D = 3\beta + 2x - x^2 + x(\beta + 1)(x - 2) \]

### Step 4: Factor Out \((x-2)(x-1)\)

We can rearrange and factor out \((x-2)(x-1)\):

\[ D = (x-2)(x-1)(\beta) \]

### Step 5: Find the Quotient

To find the quotient when \(D\) is divided by \((x-2)(x-1)\):

\[ \text{Quotient} = \beta \]

### Conclusion

Thus, we have proved that \((x-2)(x-1)\) is a factor of the determinant \(D\) and the quotient is \(\beta\).