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Let's factor the given expression completely:

\[
15y^7 - 21y^6 - 18y^5
\]

### Step 1: Find the Greatest Common Factor (GCF)
The coefficients are 15, 21, and 18. The GCF of these numbers is 3. The lowest power of \( y \) in the terms is \( y^5 \).

So, the GCF is \( 3y^5 \).

### Step 2: Factor out the GCF
We can factor \( 3y^5 \) from each term:

\[
3y^5 (5y^2 - 7y - 6)
\]

### Step 3: Factor the remaining quadratic
Now, factor the quadratic \( 5y^2 - 7y - 6 \).

We look for two numbers that multiply to \( 5 \times -6 = -30 \) and add to -7. These numbers are -10 and 3.

### Step 4: Decompose and factor
Rewrite \( -7y \) as \( -10y + 3y \):

\[
5y^2 - 10y + 3y - 6
\]

Now, factor by grouping:

\[
(5y^2 - 10y) + (3y - 6)
\]

\[
5y(y - 2) + 3(y - 2)
\]

Factor out the common term \( (y - 2) \):

\[
(5y + 3)(y - 2)
\]

### Final Factored Form
Putting it all together, we get:

\[
3y^5(5y + 3)(y - 2)
\]

This is the completely factored form of the given expression.

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you determine the GCF of terms with variables and coefficients?
2. What is the process of factoring by grouping?
3. How can you verify if a factorization is correct?
4. Why is factoring important in algebra?
5. What if the quadratic was not factorable—what methods could be used?

**Tip:** Always start by factoring out the greatest common factor (GCF) before attempting to factor further. It simplifies the process.

Learn how to factor the expression 15y^7 - 21y^6 - 18y^5 completely. This problem covers essential algebraic concepts such as finding the Greatest Common Factor (GCF) and factoring by grouping, making it suitable for students in Grades 9-10.

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