Ana multiplied [tex]\((3p - 7)\)[/tex] and [tex]\((2p^2 - 3p - 4)\)[/tex]. Her work is shown in the table below:

[tex]\[
(3p - 7)(2p^2 - 3p - 4)
\][/tex]

[tex]\[
\begin{array}{|c|c|c|c|}
\hline
& 2p^2 & -3p & -4 \\
\hline
3p & 6p^3 & -9p^2 & -12p \\
\hline
-7 & -14p^2 & 21p & 28 \\
\hline
\end{array}
\][/tex]

Which is the product?

A. [tex]\(6p^3 + 23p^2 + 9p + 28\)[/tex]
B. [tex]\(6p^3 - 23p^2 - 9p + 28\)[/tex]
C. [tex]\(6p^3 - 23p^2 + 9p + 28\)[/tex]
D. [tex]\(6p^3 + 23p^2 - 9p + 28\)[/tex]



Answer :

Let's analyze the given problem and the work done step-by-step.

We are multiplying the two polynomials [tex]\((3p - 7)\)[/tex] and [tex]\((2p^2 - 3p - 4)\)[/tex] using the table provided.

[tex]\[ (3p - 7) \left( 2p^2 - 3p - 4 \right) \][/tex]

The table given looks like:
[tex]\[ \begin{array}{|c|c|c|c|} \hline & 2p^2 & -3p & -4 \\ \hline 3p & 6p^3 & -9p^2 & -12p \\ \hline -7 & -14p^2 & 21p & 28 \\ \hline \end{array} \][/tex]

Let's interpret what this table represents:

1. Multiplying [tex]\(3p\)[/tex] with each term of the second polynomial:
[tex]\[ 3p \cdot 2p^2 = 6p^3 \][/tex]
[tex]\[ 3p \cdot (-3p) = -9p^2 \][/tex]
[tex]\[ 3p \cdot (-4) = -12p \][/tex]

2. Multiplying [tex]\(-7\)[/tex] with each term of the second polynomial:
[tex]\[ -7 \cdot 2p^2 = -14p^2 \][/tex]
[tex]\[ -7 \cdot (-3p) = 21p \][/tex]
[tex]\[ -7 \cdot (-4) = 28 \][/tex]

Now, combining the terms according to their degrees:

- [tex]\(p^3\)[/tex] term:
[tex]\[ 6p^3 \][/tex]

- [tex]\(p^2\)[/tex] terms:
[tex]\[ -9p^2 + (-14p^2) = -23p^2 \][/tex]

- [tex]\(p\)[/tex] terms:
[tex]\[ -12p + 21p = 9p \][/tex]

- Constant term:
[tex]\[ 28 \][/tex]

Thus, combining all these results, the final polynomial is:
[tex]\[ 6p^3 - 23p^2 + 9p + 28 \][/tex]

So, the correct product is:
[tex]\[ 6p^3 - 23p^2 + 9p + 28 \][/tex]

Therefore, the correct option is:
[tex]\[ \boxed{6 p^3 - 23 p^2 + 9 p + 28} \][/tex]