Question

what is the product?
(4y - 3)(2y² + 3y - 5)
○ 8y³ + 3y + 15
○ 8y³ - 23y + 15
○ 8y³ - 6y² - 17y + 15
○ 8y³ + 6y² - 29y + 15

Explanation:

Step1: Distribute $4y$ to each term

$4y \cdot 2y^2 + 4y \cdot 3y + 4y \cdot (-5)$
$= 8y^3 + 12y^2 - 20y$

Step2: Distribute $-3$ to each term

$-3 \cdot 2y^2 + (-3) \cdot 3y + (-3) \cdot (-5)$
$= -6y^2 - 9y + 15$

Step3: Combine all resulting terms

$8y^3 + 12y^2 - 20y - 6y^2 - 9y + 15$

Step4: Combine like terms

$8y^3 + (12y^2 - 6y^2) + (-20y - 9y) + 15$
$= 8y^3 + 6y^2 - 29y + 15$

Answer:

D. $8y^3 + 6y^2 - 29y + 15$