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To ensure the given function is continuous at all points, including \(x = 2\), we must find a value of \(k\) such that the limit of \(f(x)\) as \(x\) approaches 2 from both sides equals \(f(2)\), which is \(k\).

### Step 1: Simplifying the expression for \(x \neq 2\)
The given function for \(x \neq 2\) is:

\[
f(x) = \frac{5x^3 - 10x^2}{x - 2}
\]

Let's factor the numerator \(5x^3 - 10x^2\).

\[
5x^3 - 10x^2 = 5x^2(x - 2)
\]

Thus, the expression becomes:

\[
f(x) = \frac{5x^2(x - 2)}{x - 2}
\]

For \(x \neq 2\), we can cancel the \((x - 2)\) term in the numerator and denominator:

\[
f(x) = 5x^2 \quad \text{for} \quad x \neq 2
\]

### Step 2: Finding the limit as \(x \to 2\)
To ensure continuity at \(x = 2\), we need to check the limit of \(f(x)\) as \(x\) approaches 2. Since \(f(x) = 5x^2\) for \(x \neq 2\), we compute:

\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} 5x^2 = 5(2)^2 = 5 \times 4 = 20
\]

### Step 3: Setting \(k\) for continuity
For the function to be continuous at \(x = 2\), the limit as \(x\) approaches 2 must equal the function's value at \(x = 2\). Therefore, we set:

\[
k = 20
\]

Thus, the function is continuous for all \(x\) if \(k = 20\).

### Final Answer:
\[
k = 20
\]

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Would you like more details, or do you have any questions about the solution?

Here are five related questions to expand on this concept:

1. How do we define continuity of a function at a point?
2. What is the difference between continuity and differentiability?
3. How would the problem change if the expression had no common factors to cancel?
4. Can you explain why cancelling terms in rational functions is allowed for limits?
5. What would happen if \( k \) was not chosen to match the limit?

**Tip:** When solving continuity problems, always check if simplifying the function for removable discontinuities helps in finding the limit.

If possible, choose k so that the following function is continuous on any interval: f(x) = { (5x^3 - 10x^2) / (x - 2) for x ≠ 2; k for x = 2 }

Explore how to choose the value of k to ensure continuity of a piecewise function involving a rational expression. The solution involves simplifying the expression and evaluating the limit as x approaches a specific point, making it ideal for Grades 11-12 calculus students.

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