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 can follow these steps:

### Step 1: Write the polynomial
Let \( p(x) = x^2 - 3x + 2 \).

### Step 2: Factor the polynomial
We will factor the quadratic polynomial. We need to split the middle term, which is \(-3x\). We look for two numbers that multiply to \(2\) (the constant term) and add up to \(-3\) (the coefficient of \(x\)). The numbers \(-1\) and \(-2\) satisfy this condition.

Thus, we can rewrite the polynomial as:
\[ p(x) = x^2 - 2x - x + 2 \]
Now, we can group the terms:
\[ p(x) = (x^2 - 2x) + (-x + 2) \]
Factoring out the common terms gives us:
\[ p(x) = x(x - 2) - 1(x - 2) \]
Now, we can factor out \((x - 2)\):
\[ p(x) = (x - 2)(x - 1) \]

### Step 3: Find the zeroes
To find the zeroes, we set \( p(x) = 0 \):
\[ (x - 2)(x - 1) = 0 \]
This gives us two equations:
1. \( x - 2 = 0 \) → \( x = 2 \)
2. \( x - 1 = 0 \) → \( x = 1 \)

Thus, the zeroes of the polynomial are \( x = 1 \) and \( x = 2 \).

### Step 4: Verify the relation between zeroes and coefficients
The polynomial can be expressed in the standard form \( ax^2 + bx + c \), where:
- \( a = 1 \)
- \( b = -3 \)
- \( c = 2 \)

#### Sum of the zeroes
The sum of the zeroes \( (1 + 2) \) should equal \( -\frac{b}{a} \):
\[ 1 + 2 = 3 \]
\[ -\frac{b}{a} = -\frac{-3}{1} = 3 \]
Both sides are equal, so the relation holds.

#### Product of the zeroes
The product of the zeroes \( (1 \times 2) \) should equal \( \frac{c}{a} \):
\[ 1 \times 2 = 2 \]
\[ \frac{c}{a} = \frac{2}{1} = 2 \]
Again, both sides are equal, so this relation also holds.

### Conclusion
The zeroes of the polynomial \( p(x) = x^2 - 3x + 2 \) are \( x = 1 \) and \( x = 2 \). We have verified that the sum and product of the zeroes correspond to the coefficients of the polynomial.

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