Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
`x^(2)-2x-8`

Answer

To find the zeroes of the quadratic polynomial \(x^2 - 2x - 8\) and verify the relationship between the zeroes and the coefficients, we will follow these steps:

### Step 1: Write down the polynomial
We have the polynomial:
\[ f(x) = x^2 - 2x - 8 \]

### Step 2: Factor the polynomial
To find the zeroes, we need to factor the polynomial. We look for two numbers that multiply to \(-8\) (the constant term) and add to \(-2\) (the coefficient of \(x\)). The numbers that satisfy this are \(-4\) and \(2\).

So, we can rewrite the polynomial as:
\[ f(x) = x^2 - 4x + 2x - 8 \]
Now, we can group the terms:
\[ = (x^2 - 4x) + (2x - 8) \]
Factoring out the common terms gives us:
\[ = x(x - 4) + 2(x - 4) \]
Now we can factor by grouping:
\[ = (x - 4)(x + 2) \]

### Step 3: Set the factors to zero
To find the zeroes, we set each factor equal to zero:
1. \(x - 4 = 0\) → \(x = 4\)
2. \(x + 2 = 0\) → \(x = -2\)

Thus, the zeroes of the polynomial are:
\[ x = 4 \quad \text{and} \quad x = -2 \]

### Step 4: Verify the relationship between the zeroes and the coefficients
For a quadratic polynomial of the form \(ax^2 + bx + c\), the relationships are:
- Sum of the zeroes \((\alpha + \beta) = -\frac{b}{a}\)
- Product of the zeroes \((\alpha \beta) = \frac{c}{a}\)

In our polynomial:
- \(a = 1\)
- \(b = -2\)
- \(c = -8\)

Calculating the sum of the zeroes:
\[ \alpha + \beta = 4 + (-2) = 2 \]
Calculating \(-\frac{b}{a}\):
\[ -\frac{-2}{1} = 2 \]

Calculating the product of the zeroes:
\[ \alpha \beta = 4 \times (-2) = -8 \]
Calculating \(\frac{c}{a}\):
\[ \frac{-8}{1} = -8 \]

### Conclusion
Both the sum and product of the zeroes match the relationships derived from the coefficients:
- Sum of zeroes: \(2\) (matches \(-\frac{b}{a}\))
- Product of zeroes: \(-8\) (matches \(\frac{c}{a}\))

Thus, the relationships are verified.