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The given expression is:

\[
15uv^2 + 70uv + 40u
\]

To factor this expression, we first notice that there is a common factor of \( 5u \) in all the terms. We can factor that out:

\[
5u(3v^2 + 14v + 8)
\]

Now, we need to factor the quadratic expression inside the parentheses: \( 3v^2 + 14v + 8 \).

We look for two numbers that multiply to \( 3 \times 8 = 24 \) and add up to \( 14 \). These numbers are \( 12 \) and \( 2 \). So, we rewrite the middle term:

\[
3v^2 + 12v + 2v + 8
\]

Now, we factor by grouping:

\[
(3v^2 + 12v) + (2v + 8)
\]

Factor each group:

\[
3v(v + 4) + 2(v + 4)
\]

Now, factor out the common binomial factor \( (v + 4) \):

\[
(3v + 2)(v + 4)
\]

So, the fully factored form of the original expression is:

\[
5u(3v + 2)(v + 4)
\]

Would you like further details or clarifications? Here are some related questions:

1. How do we identify common factors in an expression?
2. Why do we use factoring by grouping in quadratics?
3. Can this method be used for higher degree polynomials?
4. What happens if the quadratic inside is not factorable by integers?
5. How can we verify if the factored form is correct?

**Tip:** Always double-check your factored expressions by expanding them to ensure accuracy!

Learn how to factor the algebraic expression 15uv^2 + 70uv + 40u with a step-by-step breakdown. This problem covers key concepts such as factoring by grouping and factoring quadratic expressions. Ideal for students in Grades 9-12.

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