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Answer :
To find the product of the polynomials
$$ (2x^2 + 3x - 1) \quad \text{and} \quad (3x + 5), $$
we multiply each term in the first polynomial by each term in the second polynomial.
1. Multiply the first term of the first polynomial by each term in the second polynomial:
\[
2x^2 \cdot 3x = 6x^3,
\]
\[
2x^2 \cdot 5 = 10x^2.
\]
2. Multiply the second term of the first polynomial by each term in the second polynomial:
\[
3x \cdot 3x = 9x^2,
\]
\[
3x \cdot 5 = 15x.
\]
3. Multiply the third term of the first polynomial by each term in the second polynomial:
\[
(-1) \cdot 3x = -3x,
\]
\[
(-1) \cdot 5 = -5.
\]
Now, combine like terms:
- For $x^3$: There is only one term,
\[
6x^3.
\]
- For $x^2$: Combine the terms $10x^2$ and $9x^2$,
\[
10x^2 + 9x^2 = 19x^2.
\]
- For $x$: Combine the $15x$ and $-3x$,
\[
15x - 3x = 12x.
\]
- Constant term: $-5$ remains the same.
Thus, the final product is:
$$
6x^3 + 19x^2 + 12x - 5.
$$
This corresponds to option D.
$$ (2x^2 + 3x - 1) \quad \text{and} \quad (3x + 5), $$
we multiply each term in the first polynomial by each term in the second polynomial.
1. Multiply the first term of the first polynomial by each term in the second polynomial:
\[
2x^2 \cdot 3x = 6x^3,
\]
\[
2x^2 \cdot 5 = 10x^2.
\]
2. Multiply the second term of the first polynomial by each term in the second polynomial:
\[
3x \cdot 3x = 9x^2,
\]
\[
3x \cdot 5 = 15x.
\]
3. Multiply the third term of the first polynomial by each term in the second polynomial:
\[
(-1) \cdot 3x = -3x,
\]
\[
(-1) \cdot 5 = -5.
\]
Now, combine like terms:
- For $x^3$: There is only one term,
\[
6x^3.
\]
- For $x^2$: Combine the terms $10x^2$ and $9x^2$,
\[
10x^2 + 9x^2 = 19x^2.
\]
- For $x$: Combine the $15x$ and $-3x$,
\[
15x - 3x = 12x.
\]
- Constant term: $-5$ remains the same.
Thus, the final product is:
$$
6x^3 + 19x^2 + 12x - 5.
$$
This corresponds to option D.
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