Let's find the product of the polynomials [tex]\((4y - 3)\)[/tex] and [tex]\((2y^2 + 3y - 5)\)[/tex].
Given the polynomial expression:
[tex]\[
(4y - 3)(2y^2 + 3y - 5)
\][/tex]
To find the product, we'll use the distributive property (also known as the FOIL method for binomials):
1. First, distribute [tex]\(4y\)[/tex] across the second polynomial:
[tex]\[
4y \cdot (2y^2 + 3y - 5) = 4y \cdot 2y^2 + 4y \cdot 3y + 4y \cdot (-5)
\][/tex]
Which simplifies to:
[tex]\[
8y^3 + 12y^2 - 20y
\][/tex]
2. Next, distribute [tex]\(-3\)[/tex] across the second polynomial:
[tex]\[
-3 \cdot (2y^2 + 3y - 5) = -3 \cdot 2y^2 + (-3) \cdot 3y + (-3) \cdot (-5)
\][/tex]
Which simplifies to:
[tex]\[
-6y^2 - 9y + 15
\][/tex]
3. Now, combine all the terms we obtained:
[tex]\[
(8y^3 + 12y^2 - 20y) + (-6y^2 - 9y + 15)
\][/tex]
4. Combine the like terms:
[tex]\[
8y^3 + (12y^2 - 6y^2) + (-20y - 9y) + 15
\][/tex]
Which simplifies to:
[tex]\[
8y^3 + 6y^2 - 29y + 15
\][/tex]
Therefore, the product of the polynomials [tex]\( (4y - 3) \)[/tex] and [tex]\( (2y^2 + 3y - 5) \)[/tex] is:
[tex]\[
\boxed{8y^3 + 6y^2 - 29y + 15}
\][/tex]
So, according to the given options, the correct answer is:
[tex]\[
8y^3 + 6 y^2 - 29 y + 15
\][/tex]
