- Multiply the numerical coefficients \(-8\) and \(6\) to get \(-48\).
- Apply the product rule for exponents to the \(x\) variables: \(x^5 \cdot x^2 = x^{5+2} = x^7\).
- Apply the product rule for exponents to the \(y\) variables: \(y^2 \cdot y^1 = y^{2+1} = y^3\).
- The final expression is \(\boxed{-48 x^7 y^3}\), which corresponds to option c.
Explanation
- Analysis of the Expression
To find the product of the two given expressions \(-8 x^5 y^2\) and \(6 x^2 y\), we identify the components of the product: the numerical coefficients and the variables with their exponents. The expression to evaluate is \((-8 x^5 y^2) \times (6 x^2 y)\). We recall the product rule for exponents, which states that when multiplying terms with the same base, we add their exponents: \(x^a \cdot x^b = x^{a+b}\). Additionally, we note that the variable \(y\) in the second expression is equivalent to \(y^1\).
- Calculating the Product
First, we multiply the numerical coefficients:
\[-8 \times 6 = -48.\]
Next, we combine the terms involving \(x\):
\[x^5 \times x^2 = x^{5+2} = x^7.\]
Then, we combine the terms involving \(y\):
\[y^2 \times y^1 = y^{2+1} = y^3.\]
Combining these results, we get the final expression:
\[-48 x^7 y^3.\]
- Comparing with Given Options
Comparing our result \(-48 x^7 y^3\) with the given options:
A\(-8 x^{10} y^2\)
B\(-48 x^{10} y^2\)
C\(-48 x^7 y^3\)
D\(-8 x^7 y^3\)
The calculated product matches option c. Therefore, the product of the given expression is \(\boxed{-48 x^7 y^3}\).
Examples
In physics, when calculating work done by a force over a distance, you might encounter expressions involving variables representing different physical quantities. Multiplying these monomials is a fundamental skill that allows you to simplify complex physical equations. For instance, if force is given by a function of distance \(x\) and time \(y\), multiplying such terms is a daily necessity for engineers and scientists to derive simpler, more manageable formulas.