Solve (5v+2)(4v+5) using the distributive property 

To solve \((5v + 2)(4v + 5)\) using the distributive property, we will multiply each term in the first binomial by each term in the second binomial.

Given:
\[
(5v + 2)(4v + 5)
\]

**Step 1: Distribute \(5v\) across the second binomial \((4v + 5)\):**
\[
5v \times 4v + 5v \times 5 = 20v^2 + 25v
\]

**Step 2: Distribute \(2\) across the second binomial \((4v + 5)\):**
\[
2 \times 4v + 2 \times 5 = 8v + 10
\]

**Step 3: Combine all the terms:**
\[
20v^2 + 25v + 8v + 10
\]

**Step 4: Combine the like terms (\(25v\) and \(8v\)):**
\[
20v^2 + 33v + 10
\]

So, the final answer is:
\[
\boxed{20v^2 + 33v + 10}
\]

Would you like to explore any other concepts related to this problem, or do you have any further questions?

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Here are five related questions for further practice:

1. Solve \((3x + 2)(2x + 3)\) using the distributive property.
2. Use the distributive property to expand \((6y - 4)(y + 2)\).
3. Factor the expression \(20v^2 + 33v + 10\) back into binomials.
4. Solve for \(v\) in the equation \((5v + 2)(4v + 5) = 0\).
5. Simplify \((7w - 3)(2w + 4)\) using the distributive property.

**Tip:** Breaking down the distributive property into manageable steps makes it easier to handle complex binomial products.

Learn how to solve the expression (5v + 2)(4v + 5) using the distributive property with a detailed step-by-step solution. This problem covers key algebraic concepts, including the distributive property and polynomial multiplication. Suitable for students in Grades 7-9.

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