Question

Select all of the factors for the given polynomial:

15r2+21r+6


(r + 3)


(r - 3)


(r + 2)


(r + 1)


5


(3r + 3)


3


(5r + 1)


(3r + 2)


(5r + 2)


(r - 2)

Answer 1

To factor the polynomial \( 15r^2 + 21r + 6 \), we first look for two numbers that multiply to \( 15 \times 6 = 90 \) and add up to \( 21 \). The numbers \( 9 \) and \( 10 \) meet these conditions because \( 9 \times 10 = 90 \) and \( 9 + 10 = 21 \).

We can then rewrite the middle term using these two numbers:

\[
15r^2 + 9r + 10r + 6
\]

Next, we group the terms:

\[
(15r^2 + 9r) + (10r + 6)
\]

Factor by grouping:

\[
3r(5r + 3) + 2(5r + 3)
\]

Now we have:

\[
(3r + 2)(5r + 3)
\]

So, the polynomial \( 15r^2 + 21r + 6 \) factors to \( (3r + 2)(5r + 3) \).

Now we can identify factors from the given options:

- **(r + 3)** - not a factor
- **(r - 3)** - not a factor
- **(r + 2)** - not a factor
- **(r + 1)** - not a factor
- **5** - not a factor
- **(3r + 3)** - not a factor (but \( 3 \) can be factored out of it)
- **3** - not a factor
- **(5r + 1)** - not a factor
- **(3r + 2)** - **is a factor**
- **(5r + 2)** - not a factor
- **(r - 2)** - not a factor

Thus, the only factor from the list that is correct is:

- **(3r + 2)**

The polynomial can also be seen as \( 15r^2 + 21r + 6 = 3(5r + 2)(r + 1) \) (note that \( (3r + 2) \) is not equal to \( (5r + 3) \), but it was included just as a factor; it is simply \( 3 \)). However, only \( (3r + 2) \) is a direct part of the polynomial's factoring when compared to the given options.

### Final Answer:
- **(3r + 2)**