Which is true of the function f(x)=(x-2) `(x^2 + 9)` ?

Options

A. It has no real roots
B. It has 3 real roots
C. It has 1 real root and 2 complex roots.
D. It has 3 complex roots .

Answer

To determine the nature of the roots of the function \( f(x) = (x - 2)(x^2 + 9) \), we can follow these steps:

### Step 1: Set the function equal to zero
To find the roots of the function, we set \( f(x) = 0 \):
\[ (x - 2)(x^2 + 9) = 0 \]

### Step 2: Solve each factor for zero
This equation is satisfied if either factor is equal to zero.

1. **For the first factor**:
\[ x - 2 = 0 \]
Solving this gives:
\[ x = 2 \]

2. **For the second factor**:
\[ x^2 + 9 = 0 \]
Rearranging gives:
\[ x^2 = -9 \]
Taking the square root of both sides results in:
\[ x = \pm \sqrt{-9} = \pm 3i \]

### Step 3: Identify the types of roots
From the solutions we found:
- The equation \( x - 2 = 0 \) gives us one real root, which is \( x = 2 \).
- The equation \( x^2 + 9 = 0 \) gives us two complex roots, \( x = 3i \) and \( x = -3i \).

### Conclusion
Thus, the function \( f(x) = (x - 2)(x^2 + 9) \) has:
- One real root: \( x = 2 \)
- Two complex roots: \( x = 3i \) and \( x = -3i \)

The correct answer is that the function has one real root and two complex roots.

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