Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:
`4x^(2)-4x+1`

Explanation

To find the zeros of the quadratic polynomial \(4x^2 - 4x + 1\) and verify the relationship between the zeros and the coefficients, we can follow these steps:

### Step 1: Set the polynomial equal to zero
We start by setting the polynomial equal to zero:
\[ 4x^2 - 4x + 1 = 0 \]

### Step 2: Factor the polynomial
We can factor the quadratic polynomial. We look for two numbers that multiply to \(4 \times 1 = 4\) (the product of \(A\) and \(C\)) and add up to \(-4\) (the coefficient \(B\)).

Notice that:
\[ 4x^2 - 4x + 1 = (2x - 1)(2x - 1) = (2x - 1)^2 \]

### Step 3: Solve for the zeros
Now, we set each factor equal to zero:
\[ (2x - 1) = 0 \]
Solving for \(x\):
\[ 2x = 1 \implies x = \frac{1}{2} \]

Since it is a perfect square, the zero is repeated:
Thus, the zeros of the polynomial are:
\[ x = \frac{1}{2} \quad \text{(with multiplicity 2)} \]

### Step 4: Verify the relationship between the zeros and coefficients
For a quadratic polynomial of the form \(Ax^2 + Bx + C\), the relationships between the zeros (\(\alpha\) and \(\beta\)) and the coefficients are given by:
- Sum of the zeros: \(\alpha + \beta = -\frac{B}{A}\)
- Product of the zeros: \(\alpha \cdot \beta = \frac{C}{A}\)

Here, \(A = 4\), \(B = -4\), and \(C = 1\).

#### Calculate the sum of the zeros:
\[ \alpha + \beta = \frac{1}{2} + \frac{1}{2} = 1 \]
Now, calculate \(-\frac{B}{A}\):
\[ -\frac{-4}{4} = 1 \]

#### Calculate the product of the zeros:
\[ \alpha \cdot \beta = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]
Now, calculate \(\frac{C}{A}\):
\[ \frac{1}{4} = \frac{1}{4} \]

### Conclusion
Both relationships are verified:
- Sum of the zeros: \(1\) equals \(-\frac{B}{A}\)
- Product of the zeros: \(\frac{1}{4}\) equals \(\frac{C}{A}\)

Thus, the zeros of the polynomial \(4x^2 - 4x + 1\) are \(\frac{1}{2}\) (with multiplicity 2), and the relationships between the zeros and coefficients are verified.