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: Distribute \(2p\)

Multiply \(2p\) with each term in \(3p^2 + 4p - 3\):
\(2p \times 3p^2 = 6p^3\), \(2p \times 4p = 8p^2\), \(2p \times (-3) = -6p\).

Step2: Distribute \(7\)

Multiply \(7\) with each term in \(3p^2 + 4p - 3\):
\(7 \times 3p^2 = 21p^2\), \(7 \times 4p = 28p\), \(7 \times (-3) = -21\).

Step3: Combine like terms

• For \(p^3\): \(6p^3\) (only term).
• For \(p^2\): \(8p^2 + 21p^2 = 29p^2\).
• For \(p\): \(-6p + 28p = 22p\)? Wait, no—wait, original distribution: \(2p \times (-3) = -6p\), \(7 \times 4p = 28p\). Wait, no, wait: Wait, the second polynomial is \(3p^2 + 4p - 3\). So \(2p \times (-3) = -6p\), \(7 \times 4p = 28p\). Then \( -6p + 28p = 22p\)? Wait, no, wait—wait, no, I made a mistake. Wait, the first distribution: \(2p(3p^2 + 4p - 3) = 6p^3 + 8p^2 - 6p\). Then \(7(3p^2 + 4p - 3) = 21p^2 + 28p - 21\). Now combine:

\(6p^3 + (8p^2 + 21p^2) + (-6p + 28p) + (-21)\)
\(= 6p^3 + 29p^2 + 22p - 21\)? Wait, no—wait, no, the options have \( -22p\) or \(+22p\). Wait, wait, no: Wait, \(2p \times (-3) = -6p\), \(7 \times 4p = 28p\). So \(-6p + 28p = 22p\)? But the options: Let me recheck. Wait, the problem is \((2p + 7)(3p^2 + 4p - 3)\). Let's do it again:

First, \(2p \times 3p^2 = 6p^3\)
\(2p \times 4p = 8p^2\)
\(2p \times (-3) = -6p\)

Then \(7 \times 3p^2 = 21p^2\)
\(7 \times 4p = 28p\)
\(7 \times (-3) = -21\)

Now add all terms:
\(6p^3 + 8p^2 - 6p + 21p^2 + 28p - 21\)

Combine like terms:

• \(p^3\): \(6p^3\)
• \(p^2\): \(8p^2 + 21p^2 = 29p^2\)
• \(p\): \(-6p + 28p = 22p\)
• constant: \(-21\)

Wait, but the options: Let's check the options. The second option is \(6p^3 + 29p^2 - 22p + 21\), third is \(6p^3 + 29p^2 + 22p - 21\). Wait, I must have messed up the signs. Wait, \(7 \times (-3) = -21\), correct. \(2p \times (-3) = -6p\), \(7 \times 4p = 28p\). So \(-6p + 28p = 22p\), and constant is \(-21\). So the product is \(6p^3 + 29p^2 + 22p - 21\), which is the third option? Wait, no, wait the options:

Option 1: \(6p^3 + 29p^2 - 34p + 21\)
Option 2: \(6p^3 + 29p^2 - 22p + 21\)
Option 3: \(6p^3 + 29p^2 + 22p - 21\)
Option 4: \(6p^3 + 29p^2 + 34p - 21\)

Wait, so my calculation gives \(6p^3 + 29p^2 + 22p - 21\), which is option 3? Wait, but let me check again. Wait, maybe I flipped a sign. Let's use the distributive property (FOIL for trinomials):

\((2p + 7)(3p^2 + 4p - 3) = 2p(3p^2) + 2p(4p) + 2p(-3) + 7(3p^2) + 7(4p) + 7(-3)\)
\(= 6p^3 + 8p^2 - 6p + 21p^2 + 28p - 21\)

Now combine like terms:

• \(p^3\): \(6p^3\)
• \(p^2\): \(8p^2 + 21p^2 = 29p^2\)
• \(p\): \(-6p + 28p = 22p\)
• constant: \(-21\)

So the product is \(6p^3 + 29p^2 + 22p - 21\), which is option 3: \(6p^3 + 29p^2 + 22p - 21\). Wait, but let me check the signs again. The constant term: \(7 \times (-3) = -21\), correct. The \(p\) term: \(-6p + 28p = 22p\), correct. So yes, option 3.

Answer:

C. \(6p^3 + 29p^2 + 22p - 21\) (assuming the options are labeled A, B, C, D with C being the third option as per the order: first option is A, second B, third C, fourth D)