Question

the table has two columns with headers (-x) and (-2), and three rows with headers (-2x^2), (-6x), and (4). each cell at the intersection of a row and column is a blank square to be filled.

Explanation:

To fill in the table, we need to multiply the terms in the rows by the terms in the columns. Let's go step by step:

Step 1: Multiply \(-2x^2\) with \(-x\)

To find the product of \(-2x^2\) and \(-x\), we use the rule of exponents \(a^m \cdot a^n = a^{m + n}\) and the rule of signs (negative times negative is positive).

$$ (-2x^2) \cdot (-x) = (-2) \cdot (-1) \cdot x^{2 + 1} = 2x^3 $$

Step 2: Multiply \(-2x^2\) with \(-2\)

Using the rule of signs (negative times negative is positive) and multiplying the coefficients and constants.

$$ (-2x^2) \cdot (-2) = (-2) \cdot (-2) \cdot x^2 = 4x^2 $$

Step 3: Multiply \(-6x\) with \(-x\)

Using the rule of exponents and signs (negative times negative is positive).

$$ (-6x) \cdot (-x) = (-6) \cdot (-1) \cdot x^{1 + 1} = 6x^2 $$

Step 4: Multiply \(-6x\) with \(-2\)

Using the rule of signs (negative times negative is positive) and multiplying the coefficients and variables.

$$ (-6x) \cdot (-2) = (-6) \cdot (-2) \cdot x = 12x $$

Step 5: Multiply \(4\) with \(-x\)

Using the rule of signs (positive times negative is negative).

$$ 4 \cdot (-x) = -4x $$

Step 6: Multiply \(4\) with \(-2\)

Using the rule of signs (positive times negative is negative).

$$ 4 \cdot (-2) = -8 $$

Now, let's fill in the table:

	\(-x\)	\(-2\)
\(-6x\)	\(6x^2\)	\(12x\)
\(4\)	\(-4x\)	\(-8\)

Answer:

The filled table is as follows:

	\(-x\)	\(-2\)
\(-6x\)	\(6x^2\)	\(12x\)
\(4\)	\(-4x\)	\(-8\)