Find zeroes of the polynomial p(x)=`x^(3)-9x^(2)+26x-24`, if it is given that the product of its two zeroes is 8.

Explanation

To find the zeroes of the polynomial \( p(x) = x^3 - 9x^2 + 26x - 24 \), given that the product of two of its zeroes is 8, we can follow these steps:

### Step 1: Identify the relationships between the zeroes
Let the zeroes of the polynomial be \( \alpha, \beta, \) and \( \gamma \). We know:
- The sum of the zeroes \( \alpha + \beta + \gamma = -\frac{b}{a} = 9 \) (where \( b = -9 \) and \( a = 1 \)).
- The sum of the product of the zeroes taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = 26 \) (where \( c = 26 \)).
- The product of the zeroes \( \alpha\beta\gamma = -\frac{d}{a} = 24 \) (where \( d = -24 \)).

### Step 2: Use the given information
We are given that \( \alpha \beta = 8 \). We can use this information to find \( \gamma \):
\[ \alpha \beta \gamma = 24 \implies 8 \gamma = 24 \implies \gamma = \frac{24}{8} = 3. \]

### Step 3: Find \( \alpha + \beta \)
Now, we can use the sum of the zeroes:
\[ \alpha + \beta + \gamma = 9 \implies \alpha + \beta + 3 = 9 \implies \alpha + \beta = 9 - 3 = 6. \]

### Step 4: Set up equations for \( \alpha \) and \( \beta \)
Now we have two equations:
1. \( \alpha + \beta = 6 \) (Equation 1)
2. \( \alpha \beta = 8 \) (Equation 2)

### Step 5: Solve for \( \alpha \) and \( \beta \)
We can express \( \beta \) in terms of \( \alpha \) using Equation 1:
\[ \beta = 6 - \alpha. \]
Substituting this into Equation 2:
\[ \alpha(6 - \alpha) = 8. \]
Expanding this gives:
\[ 6\alpha - \alpha^2 = 8 \implies \alpha^2 - 6\alpha + 8 = 0. \]

### Step 6: Solve the quadratic equation
Now we can solve the quadratic equation \( \alpha^2 - 6\alpha + 8 = 0 \) using the quadratic formula:
\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm \sqrt{4}}{2} = \frac{6 \pm 2}{2}. \]
This gives us:
\[ \alpha = \frac{8}{2} = 4 \quad \text{or} \quad \alpha = \frac{4}{2} = 2. \]

### Step 7: Find \( \beta \)
Using \( \alpha + \beta = 6 \):
- If \( \alpha = 4 \), then \( \beta = 6 - 4 = 2 \).
- If \( \alpha = 2 \), then \( \beta = 6 - 2 = 4 \).

### Final Step: List all zeroes
Thus, the zeroes of the polynomial are:
\[ \alpha = 4, \beta = 2, \gamma = 3. \]
So, the zeroes of the polynomial \( p(x) = x^3 - 9x^2 + 26x - 24 \) are \( 2, 3, \) and \( 4 \).