Find zeroes of the polynormial `x^(2)-3x+2` and verify the relation between its zeroes and coefficients.

Explanation

To find the zeroes of the polynomial \( P(x) = x^2 - 3x + 2 \) and verify the relation between its zeroes and coefficients, we will follow these steps:

### Step 1: Identify the polynomial
The given polynomial is:
\[ P(x) = x^2 - 3x + 2 \]

### Step 2: Set the polynomial equal to zero
To find the zeroes, we set the polynomial equal to zero:
\[ x^2 - 3x + 2 = 0 \]

### Step 3: Factor the polynomial
We need to factor the quadratic equation. We look for two numbers that multiply to \( 2 \) (the constant term) and add up to \( -3 \) (the coefficient of \( x \)). The numbers that satisfy these conditions are \( -1 \) and \( -2 \).

Thus, we can factor the polynomial as:
\[ (x - 1)(x - 2) = 0 \]

### Step 4: Find the zeroes
Setting each factor equal to zero gives us the zeroes:
1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( x - 2 = 0 \) → \( x = 2 \)

So, the zeroes of the polynomial are:
\[ x = 1 \quad \text{and} \quad x = 2 \]

### Step 5: Verify the relation between zeroes and coefficients
For a quadratic polynomial of the form \( ax^2 + bx + c \):
- The sum of the zeroes \( (r_1 + r_2) \) is given by \( -\frac{b}{a} \)
- The product of the zeroes \( (r_1 \cdot r_2) \) is given by \( \frac{c}{a} \)

In our polynomial:
- \( a = 1 \)
- \( b = -3 \)
- \( c = 2 \)

**Sum of the zeroes:**
\[ r_1 + r_2 = 1 + 2 = 3 \]
Calculating using the coefficients:
\[ -\frac{b}{a} = -\frac{-3}{1} = 3 \]

**Product of the zeroes:**
\[ r_1 \cdot r_2 = 1 \cdot 2 = 2 \]
Calculating using the coefficients:
\[ \frac{c}{a} = \frac{2}{1} = 2 \]

### Conclusion
Both the sum and product of the zeroes match the relations derived from the coefficients:
- Sum of zeroes \( = 3 \)
- Product of zeroes \( = 2 \)

Thus, we have verified the relation between the zeroes and coefficients.

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