To expand the expression \((y-4)(2y+3)\), we will again use the distributive property (or the FOIL method) to multiply each term in the first binomial by each term in the second binomial.

Here’s how we can expand it step-by-step:

\[
(y-4)(2y+3) = y \cdot 2y + y \cdot 3 + (-4) \cdot 2y + (-4) \cdot 3
\]

Now, let's calculate each term:

1. **First**: \(y \cdot 2y = 2y^2\)
2. **Outer**: \(y \cdot 3 = 3y\)
3. **Inner**: \((-4) \cdot 2y = -8y\)
4. **Last**: \((-4) \cdot 3 = -12\)

Now, combine all these terms:

\[
2y^2 + 3y - 8y - 12
\]

Combine like terms \((3y - 8y)\):

\[
2y^2 - 5y - 12
\]

So, the expanded form of \((y-4)(2y+3)\) is:

\[
2y^2 - 5y - 12
\]

Would you like more details or have any questions?

Here are some related questions to explore:

1. How do you factor the quadratic expression \(2y^2 - 5y - 12\)?
2. What are the roots of the equation \(2y^2 - 5y - 12 = 0\)?
3. What changes in the expansion if the binomials are \((y + 4)(2y - 3)\)?
4. How does the distributive property apply when multiplying binomials?
5. How would you graph the quadratic function \(f(y) = 2y^2 - 5y - 12\)?

**Tip**: When expanding expressions, it helps to write down each step to avoid mistakes, especially when dealing with negative signs.

Learn how to expand the expression (y-4)(2y+3) using the FOIL method with a detailed step-by-step solution. This problem covers algebraic concepts such as polynomials and the distributive property. Suitable for students in Grades 7-9.

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