Sure! Let's fill out the given table by multiplying each row heading by the column headings step-by-step.
First, we will set up a table to organize our products:
[tex]\[
\begin{array}{|c|c|c|}
\hline
& 5t & -4 \\
\hline
-16t & & \\
\hline
4t & & \\
\hline
5 & & \\
\hline
\end{array}
\][/tex]
Now we will go row by row to calculate the products:
### First row:
- For the first cell: multiply \(-16t\) by \(5t\) to get \(\left(-16t \cdot 5t\right) = -80t^2\).
- For the second cell: multiply \(-16t\) by \(-4\) to get \(\left(-16t \cdot (-4)\right) = 64t\).
So the table now looks like this:
[tex]\[
\begin{array}{|c|c|c|}
\hline
& 5t & -4 \\
\hline
-16t & -80t^2 & 64t \\
\hline
4t & & \\
\hline
5 & & \\
\hline
\end{array}
\][/tex]
### Second row:
- For the first cell: multiply \(4t\) by \(5t\) to get \(\left(4t \cdot 5t\right) = 20t^2\).
- For the second cell: multiply \(4t\) by \(-4\) to get \(\left(4t \cdot (-4)\right) = -16t\).
So the table now looks like this:
[tex]\[
\begin{array}{|c|c|c|}
\hline
& 5t & -4 \\
\hline
-16t & -80t^2 & 64t \\
\hline
4t & 20t^2 & -16t \\
\hline
5 & & \\
\hline
\end{array}
\][/tex]
### Third row:
- For the first cell: multiply \(5\) by \(5t\) to get \(\left(5 \cdot 5t\right) = 25t\).
- For the second cell: multiply \(5\) by \(-4\) to get \(\left(5 \cdot (-4)\right) = -20\).
So the final table looks like this:
[tex]\[
\begin{array}{|c|c|c|}
\hline
& 5t & -4 \\
\hline
-16t & -80t^2 & 64t \\
\hline
4t & 20t^2 & -16t \\
\hline
5 & 25t & -20 \\
\hline
\end{array}
\][/tex]
This is our completed table with all the products calculated!
