How many zeroes will be there for the polynomial f(x) = `(x- 2)^2` + 4?

Options

A. 0
B. 1
C. 2
D. 3

Answer

To determine how many zeros the polynomial \( f(x) = (x - 2)^2 + 4 \) has, we can follow these steps:

### Step 1: Rewrite the Polynomial
Start with the given polynomial:
\[ f(x) = (x - 2)^2 + 4 \]

### Step 2: Analyze the Expression
The expression \( (x - 2)^2 \) is always non-negative because it is a square term. The minimum value of \( (x - 2)^2 \) occurs when \( x = 2 \), which gives:
\[ (x - 2)^2 = 0 \]
Thus, we have:
\[ f(2) = 0 + 4 = 4 \]
This means the minimum value of \( f(x) \) is 4.

### Step 3: Determine the Zeros
Since \( f(x) \) achieves a minimum value of 4 and does not go below this value, the polynomial \( f(x) \) does not cross the x-axis. Therefore, it has no real zeros.

### Step 4: Conclusion
Since the polynomial does not have any points where \( f(x) = 0 \), we conclude that:
\[ \text{The number of zeros of the polynomial } f(x) = (x - 2)^2 + 4 \text{ is } 0. \]

### Final Answer
The polynomial has **0 zeros**.

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