Question

what is the product of (2p + 7)(3p² + 4p − 3)?
○ 6p³ + 29p² − 34p + 21
○ 6p³ + 29p² − 22p + 21
○ 6p³ + 29p² + 22p − 21
○ 6p³ + 29p² + 34p − 21

Explanation:

Step1: Apply distributive property (FOIL for binomial and trinomial)

Multiply \(2p\) by each term in \(3p^2 + 4p - 3\) and \(7\) by each term in \(3p^2 + 4p - 3\):
\(2p(3p^2)+2p(4p)+2p(-3)+7(3p^2)+7(4p)+7(-3)\)

Step2: Simplify each product

\(6p^3 + 8p^2 - 6p + 21p^2 + 28p - 21\)

Step3: Combine like terms

Combine \(p^2\) terms: \(8p^2 + 21p^2 = 29p^2\)
Combine \(p\) terms: \(-6p + 28p = 22p\)
So the expression becomes \(6p^3 + 29p^2 + 22p - 21\)

Answer:

\(6p^3 + 29p^2 + 22p - 21\) (corresponding to the third option: \(6p^3 + 29p^2 + 22p - 21\))