Question

ilana multiplied 3p - 7 and 2p² - 3p - 4. her work is shown in the table.
(3p - 7)(2p² - 3p - 4)
which is the product?
○ 6p³ + 23p² + 9p + 28
○ 6p³ - 23p² - 9p + 28
○ 6p³ - 23p² + 9p + 28
○ 6p³ + 23p² - 9p + 28

Explanation:

Step1: Identify the multiplication

We need to find the product of \((3p - 7)(2p^2 - 3p - 4)\) by adding all the terms in the table.

Step2: List all terms from the table

The terms from the table are: \(6p^3\) (from \(3p\times2p^2\)), \(-9p^2\) (from \(3p\times(-3p)\)), \(-12p\) (from \(3p\times(-4)\)), \(-14p^2\) (from \(-7\times2p^2\)), \(21p\) (from \(-7\times(-3p)\)), and \(28\) (from \(-7\times(-4)\)).

Step3: Combine like terms

• For \(p^3\) term: \(6p^3\)
• For \(p^2\) terms: \(-9p^2-14p^2=-23p^2\)
• For \(p\) terms: \(-12p + 21p=9p\)
• Constant term: \(28\)

Step4: Write the final polynomial

Combining all terms, we get \(6p^3-23p^2 + 9p+28\) (Wait, wait, no, wait. Wait, let's re - check the table. Wait, the first row of the table: the first column is \(2p^2\), \(3p\), \(-4\) (wait, no, the leftmost column is \(3p\), \(-7\)? Wait, no, the table is a multiplication table for \((3p - 7)(2p^2-3p - 4)\). So the rows are \(3p\) and \(-7\), columns are \(2p^2\), \(-3p\), \(-4\).

So:

• \(3p\times2p^2 = 6p^3\)
• \(3p\times(-3p)=-9p^2\)
• \(3p\times(-4)=-12p\)
• \(-7\times2p^2=-14p^2\)
• \(-7\times(-3p) = 21p\)
• \(-7\times(-4)=28\)

Now, combine like terms:

• \(p^3\): \(6p^3\)
• \(p^2\): \(-9p^2-14p^2=-23p^2\)
• \(p\): \(-12p + 21p = 9p\)
• Constant: \(28\)

So the product is \(6p^3-23p^2 + 9p + 28\)? Wait, but looking at the options, the first option is \(6p^{3}+23p^{2}+9p + 28\), second is \(6p^{3}-23p^{2}-9p + 28\), third is \(6p^{3}-23p^{2}+9p + 28\), fourth is \(6p^{3}+23p^{2}-9p + 28\). Wait, I think I made a mistake in the sign of the \(p^2\) term from \(-7\times2p^2\). Wait, \(-7\times2p^2=-14p^2\) and \(3p\times(-3p)=-9p^2\), so \(-9p^2-14p^2=-23p^2\) (correct). For the \(p\) terms: \(3p\times(-4)=-12p\) and \(-7\times(-3p) = 21p\), so \(-12p + 21p = 9p\) (correct). So the polynomial is \(6p^3-23p^2+9p + 28\), which is the third option. Wait, the third option is \(6p^{3}-23p^{2}+9p + 28\) (the original options: third option is \(6p^{3}-23p^{2}+9p + 28\)).

Wait, let's re - express the multiplication:

\((3p-7)(2p^{2}-3p - 4)=3p\times(2p^{2}-3p - 4)-7\times(2p^{2}-3p - 4)\)

\(=3p\times2p^{2}-3p\times3p-3p\times4-7\times2p^{2}+7\times3p + 7\times4\)

\(=6p^{3}-9p^{2}-12p-14p^{2}+21p + 28\)

Now, combine like terms:

• \(p^{3}\): \(6p^{3}\)
• \(p^{2}\): \(-9p^{2}-14p^{2}=-23p^{2}\)
• \(p\): \(-12p + 21p = 9p\)
• Constants: \(28\)

So the product is \(6p^{3}-23p^{2}+9p + 28\)

Answer:

The third option: \(6p^{3}-23p^{2}+9p + 28\) (assuming the third option is \(6p^{3}-23p^{2}+9p + 28\) as per the given options)