From question number 1 to 16, find zeroes of the given quadratic polynomials and verify the relation between zeroes and coefficients :
`x^(2)+4x-5`

Explanation

To find the zeroes of the quadratic polynomial \( x^2 + 4x - 5 \) and verify the relationship between the zeroes and coefficients, we can follow these steps:

### Step 1: Identify the coefficients
The given polynomial is \( x^2 + 4x - 5 \). We can compare this with the standard form of a quadratic polynomial \( ax^2 + bx + c \).

- Here, \( a = 1 \), \( b = 4 \), and \( c = -5 \).

### Step 2: Find the sum and product of the roots
Using the relationships:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

Substituting the values:
- Sum of the roots:
\[ \alpha + \beta = -\frac{4}{1} = -4 \]
- Product of the roots:
\[ \alpha \beta = \frac{-5}{1} = -5 \]

### Step 3: Factor the polynomial
To find the roots, we can factor the polynomial \( x^2 + 4x - 5 \). We look for two numbers that multiply to \( -5 \) (the product) and add to \( 4 \) (the sum).

The numbers \( 5 \) and \( -1 \) satisfy these conditions:
- \( 5 \times (-1) = -5 \)
- \( 5 + (-1) = 4 \)

Thus, we can rewrite the polynomial as:
\[ x^2 + 5x - x - 5 = 0 \]
Grouping the terms:
\[ (x^2 + 5x) + (-x - 5) = 0 \]
Factoring by grouping:
\[ x(x + 5) - 1(x + 5) = 0 \]
Factoring out \( (x + 5) \):
\[ (x - 1)(x + 5) = 0 \]

### Step 4: Solve for the roots
Setting each factor to zero gives us the roots:
1. \( x - 1 = 0 \) → \( x = 1 \) (first root, \( \alpha \))
2. \( x + 5 = 0 \) → \( x = -5 \) (second root, \( \beta \))

### Step 5: Verify the relationships
Now we verify the relationships:
- Sum of the roots:
\[ \alpha + \beta = 1 + (-5) = -4 \]
- Product of the roots:
\[ \alpha \beta = 1 \times (-5) = -5 \]

Both relationships are verified:
- \( \alpha + \beta = -\frac{b}{a} = -4 \)
- \( \alpha \beta = \frac{c}{a} = -5 \)

### Final Result
The zeroes of the polynomial \( x^2 + 4x - 5 \) are \( 1 \) and \( -5 \).

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