To factor the polynomial \(40r^3 + 200r^2 - 5r - 25\) by grouping, we proceed as follows:

### 1. **Group the terms:**
Group the polynomial into two pairs:

\[
(40r^3 + 200r^2) - (5r + 25)
\]

### 2. **Factor out the common factors from each group:**

- **First group:** \(40r^3 + 200r^2\)
  
  The common factor is \(40r^2\), so:

  \[
  40r^2(r + 5)
  \]

- **Second group:** \(-5r - 25\)
  
  The common factor is \(-5\), so:

  \[
  -5(r + 5)
  \]

### 3. **Factor out the common binomial factor:**

Now, the expression looks like this:

\[
40r^2(r + 5) - 5(r + 5)
\]

Both terms contain the common factor \((r + 5)\). Factor this out:

\[
(r + 5)(40r^2 - 5)
\]

### 4. **Simplify further (if possible):**

Notice that \(40r^2 - 5\) has a common factor of 5, so we can factor it out:

\[
(r + 5) \cdot 5(8r^2 - 1)
\]

Since \(8r^2 - 1\) is a difference of squares, it can be further factored:

\[
8r^2 - 1 = (2\sqrt{2}r - 1)(2\sqrt{2}r + 1)
\]

### 5. **Final factorization:**

So, the completely factored form is:

\[
\boxed{5(r + 5)(2\sqrt{2}r - 1)(2\sqrt{2}r + 1)}
\]

Would you like to explore any specific part of this factorization or ask any questions?

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Here are five related questions you might explore:

1. How do you identify when to use grouping for factoring?
2. What are some examples of polynomials where grouping is not effective?
3. How do you recognize the difference of squares in a polynomial?
4. Can grouping be used to factor polynomials with more than four terms?
5. What strategies can you use when a polynomial resists easy factorization?

**Tip:** After grouping, always check if the remaining binomials can be factored further.

Learn how to factor the polynomial 40r^3 + 200r^2 - 5r - 25 step-by-step using the grouping method in algebra. This problem involves polynomial factoring with detailed explanations and includes the application of the difference of squares. Suitable for high school students.

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