Question

Find the product. Simplify your answer.
(3v+1)(4v^2+3v–3)

Answer 1

To find the product of \((3v + 1)(4v^2 + 3v - 3)\), we can use the distributive property (also known as the FOIL method for binomials).

We will distribute each term in the first polynomial \( (3v + 1) \) to each term in the second polynomial \( (4v^2 + 3v - 3) \).

1. First, distribute \(3v\):

\[
3v \cdot 4v^2 = 12v^3
\]
\[
3v \cdot 3v = 9v^2
\]
\[
3v \cdot (-3) = -9v
\]

So, the contributions from \(3v\) are:

\[
12v^3 + 9v^2 - 9v
\]

2. Next, distribute \(1\):

\[
1 \cdot 4v^2 = 4v^2
\]
\[
1 \cdot 3v = 3v
\]
\[
1 \cdot (-3) = -3
\]

So, the contributions from \(1\) are:

\[
4v^2 + 3v - 3
\]

3. Now, we combine all the results:

\[
12v^3 + 9v^2 - 9v + 4v^2 + 3v - 3
\]

4. Combine like terms:

- The \(v^3\) term: \(12v^3\)
- The \(v^2\) terms: \(9v^2 + 4v^2 = 13v^2\)
- The \(v\) terms: \(-9v + 3v = -6v\)
- The constant term: \(-3\)

Putting it all together, we get:

\[
12v^3 + 13v^2 - 6v - 3
\]

Thus, the simplified product is

\[
\boxed{12v^3 + 13v^2 - 6v - 3}
\]