If `f(x)={((x^(3)+x^(2)-16x+20)/((x-2)^(2))",",x ne 2),(k",", x =2):}` is continuous at x = 2, then the value of k is

Options

A. 1
B. 3
C. 6
D. 7

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 2 \). This means that the limit of \( f(x) \) as \( x \) approaches 2 must equal \( f(2) \).

Given:
\[ f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2}, & x \neq 2 \\ k, & x = 2 \end{cases} \]

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 2.
We need to calculate:
\[ \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2} \]

### Step 2: Substitute \( x = 2 \) into the numerator.
Calculating the numerator:
\[ 2^3 + 2^2 - 16 \cdot 2 + 20 = 8 + 4 - 32 + 20 = 0 \]
The numerator equals 0 when \( x = 2 \).

### Step 3: Factor the numerator.
Since the numerator equals 0 at \( x = 2 \), we can factor it. We can use synthetic division or polynomial long division to factor \( x^3 + x^2 - 16x + 20 \) by \( (x-2) \).

Performing synthetic division:
- Coefficients: \( 1, 1, -16, 20 \)
- Using \( 2 \):
```
2 | 1 1 -16 20
| 2 6 -20
---------------------
1 3 -10 0
```
The result is \( x^2 + 3x - 10 \). Thus,
\[ x^3 + x^2 - 16x + 20 = (x-2)(x^2 + 3x - 10) \]

### Step 4: Factor \( x^2 + 3x - 10 \).
Now we can factor \( x^2 + 3x - 10 \):
\[ x^2 + 3x - 10 = (x + 5)(x - 2) \]
Thus,
\[ x^3 + x^2 - 16x + 20 = (x - 2)^2(x + 5) \]

### Step 5: Substitute the factor back into the limit.
Now we can rewrite the function:
\[ f(x) = \frac{(x - 2)^2(x + 5)}{(x - 2)^2} = x + 5 \quad \text{for } x \neq 2 \]

### Step 6: Calculate the limit.
Now we can find the limit:
\[ \lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 5) = 2 + 5 = 7 \]

### Step 7: Set the limit equal to \( f(2) \).
For \( f(x) \) to be continuous at \( x = 2 \):
\[ f(2) = k \]
Thus, we set:
\[ k = 7 \]

### Conclusion
The value of \( k \) that makes \( f(x) \) continuous at \( x = 2 \) is:
\[ \boxed{7} \]