Find k so that `x^2-3x+k` is a factor of polynomial `x^3-6x^2+11x-6`. Also find all the zeroes of the two polynomial.

Answer

To solve the problem, we need to find the value of \( k \) such that the polynomial \( x^2 - 3x + k \) is a factor of the polynomial \( x^3 - 6x^2 + 11x - 6 \). We will also find all the zeroes of both polynomials.

### Step 1: Set up the division
Since \( x^2 - 3x + k \) is a factor of \( x^3 - 6x^2 + 11x - 6 \), we can perform polynomial long division.

### Step 2: Perform polynomial long division
1. Divide the leading term of \( x^3 \) by the leading term of \( x^2 \):
\[ x^3 \div x^2 = x \]
2. Multiply \( x \) by \( x^2 - 3x + k \):
\[ x(x^2 - 3x + k) = x^3 - 3x^2 + kx \]
3. Subtract this from \( x^3 - 6x^2 + 11x - 6 \):
\[ (x^3 - 6x^2 + 11x - 6) - (x^3 - 3x^2 + kx) = (-6x^2 + 3x^2) + (11x - kx) - 6 = -3x^2 + (11 - k)x - 6 \]

### Step 3: Continue the division
1. Now, divide the leading term of \( -3x^2 \) by the leading term of \( x^2 \):
\[ -3x^2 \div x^2 = -3 \]
2. Multiply \( -3 \) by \( x^2 - 3x + k \):
\[ -3(x^2 - 3x + k) = -3x^2 + 9x - 3k \]
3. Subtract this from \( -3x^2 + (11 - k)x - 6 \):
\[ (-3x^2 + (11 - k)x - 6) - (-3x^2 + 9x - 3k) = (11 - k - 9)x + (-6 + 3k) = (2 - k)x + (3k - 6) \]

### Step 4: Set the remainder to zero
For \( x^2 - 3x + k \) to be a factor, the remainder must be zero:
1. Set the coefficients of \( x \) and the constant term to zero:
\[ 2 - k = 0 \quad \text{(1)} \]
\[ 3k - 6 = 0 \quad \text{(2)} \]

### Step 5: Solve for \( k \)
From equation (1):
\[ k = 2 \]

From equation (2):
\[ 3k = 6 \implies k = 2 \]

Thus, \( k = 2 \).

### Step 6: Factor the polynomial
Now that we have \( k = 2 \), we can rewrite the polynomial:
\[ x^3 - 6x^2 + 11x - 6 = (x^2 - 3x + 2)(x - 3) \]

### Step 7: Factor \( x^2 - 3x + 2 \)
We can factor \( x^2 - 3x + 2 \):
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

### Step 8: Write the complete factorization
Thus, the complete factorization of the polynomial is:
\[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3) \]

### Step 9: Find the zeroes
Setting each factor to zero gives us the zeroes:
1. \( x - 1 = 0 \implies x = 1 \)
2. \( x - 2 = 0 \implies x = 2 \)
3. \( x - 3 = 0 \implies x = 3 \)

### Final Answer
- The value of \( k \) is \( 2 \).
- The zeroes of the polynomial \( x^3 - 6x^2 + 11x - 6 \) are \( 1, 2, 3 \).