Verified Answer :

  • Identify the expression as the product of two monomials: \((-8x^5y^2) \cdot (6x^2y)\).
  • Multiply the coefficients: \(-8 \cdot 6 = -48\).
  • Use exponent rules (\(x^a \cdot x^b = x^{a+b}\)) to combine variables: \(x^5 \cdot x^2 = x^7\) and \(y^2 \cdot y^1 = y^3\).
  • Conclude that the final expression is \(-48x^7y^3\), which matches option c.
  • The final answer is \(\boxed{c}\).

Explanation

  1. Analyzing the Multiplication of Monomials

To find the product of two monomials, \(-8x^5y^2\) and \(6x^2y\), we apply the rules for multiplying powers with the same base and multiply the coefficients together. The product can be written as:
\[(-8x^5y^2) \cdot (6x^2y)\]
This expression is organized by grouping the numerical coefficients and the variables with the same bases:
\[(-8 \cdot 6) \cdot (x^5 \cdot x^2) \cdot (y^2 \cdot y^1)\]
Note that \(y\) is equivalent to \(y^1\).

  1. Performing the Calculation and Identifying the Answer

Next, we perform the multiplication for each part identified in the previous step:

  1. Multiply the numerical coefficients: \(-8 \cdot 6 = -48\).
  2. Multiply the variables with the same base \(x\): using the exponent property \(x^a \cdot x^b = x^{a+b}\), we get \(x^5 \cdot x^2 = x^{5+2} = x^7\).
  3. Multiply the variables with the same base \(y\): using the same property, we get \(y^2 \cdot y^1 = y^{2+1} = y^3\).

Combining these results, the product is \(-48x^7y^3\). Comparing this to the given options:

A\(-8 x^{10} y^2\)
B\(-48 x^{10} y^2\)
C\(-48 x^7 y^3\)
D\(-8 x^7 y^3\)

We see that the correct option is c.

Examples

Understanding how to multiply monomials is a fundamental skill in algebra, useful for simplifying complex expressions and solving polynomial equations. For example, if you are designing a rectangular floor space where the length is \(6x^2y\) meters and the width is \(-8x^5y^2\) meters (assuming these expressions represent scaling factors for different material costs based on variables \(x\) and \(y\)), the total cost function (or area/volume) would involve multiplying these two expressions. Mastering these exponent rules allows you to handle such algebraic models efficiently in fields like engineering, economics, or data analysis.

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