To factor the expression \(-2x(2x^2 + 11x + 5)\), follow these steps:

### Step 1: Factor the quadratic expression inside the parentheses.
The quadratic expression is \(2x^2 + 11x + 5\). We need to factor this expression.

We are looking for two numbers that multiply to \(2 \times 5 = 10\) (the product of the coefficient of \(x^2\) and the constant term) and add to \(11\) (the coefficient of \(x\)).

These two numbers are \(10\) and \(1\).

So, we can rewrite the quadratic expression as:
\[
2x^2 + 10x + x + 5
\]

### Step 2: Factor by grouping.
Now, group the terms in pairs:
\[
(2x^2 + 10x) + (x + 5)
\]

Factor out the greatest common factor from each group:
\[
2x(x + 5) + 1(x + 5)
\]

Since both groups contain \((x + 5)\), factor it out:
\[
(x + 5)(2x + 1)
\]

### Step 3: Incorporate the original factor.
Don't forget the \(-2x\) that we factored out initially:
\[
-2x(2x^2 + 11x + 5) = -2x(x + 5)(2x + 1)
\]

So, the fully factored form of the expression is:
\[
\boxed{-2x(x + 5)(2x + 1)}
\]

Would you like more details on any of these steps, or do you have any questions?

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Here are five related questions to expand your understanding:

1. How do you factor a quadratic expression in general?
2. What is the significance of the greatest common factor in factoring?
3. Can all quadratic expressions be factored over the integers?
4. How would you factor \(x^2 - 9\) using the difference of squares method?
5. What is the process to factor a cubic polynomial?

**Tip:** Always check your factors by expanding them to ensure they multiply back to the original expression.

Understand how to factor the expression -2x(2x^2 + 11x + 5) step-by-step. This problem covers algebraic concepts, including factoring by grouping and applying the Zero Product Property, suitable for students in Grades 9-12.

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