- Factor the numerator \(4k+2\) as \(2(2k+1)\) and the denominator \(k^2-4\) as \((k-2)(k+2)\).
- Write the product as \(\frac{2(2k+1)(k-2)}{(k-2)(k+2)(2k+1)}\).
- Cancel the common factors \((2k+1)\) and \((k-2)\) from the numerator and denominator.
- Conclude that the simplified product is \(\boxed{\frac{2}{k+2}}\).
Explanation
- Analysis and Factoring
The problem asks us to simplify the algebraic expression \(\frac{4k+2}{k^2-4} \cdot \frac{k-2}{2k+1}\). We start by analyzing the components of the expression. The numerator of the first fraction is \(4k+2\), which can be factored by pulling out the common factor of \(2\), giving \(2(2k+1)\). The denominator of the first fraction, \(k^2-4\), is a difference of squares that factors into \((k-2)(k+2)\).
- Simplifying the Expression
Now we substitute these factored forms back into the original expression:
\[\frac{2(2k+1)}{(k-2)(k+2)} \cdot \frac{k-2}{2k+1}\]
By multiplying the two fractions, we combine them into a single expression:
\[\frac{2(2k+1)(k-2)}{(k-2)(k+2)(2k+1)}\]
Assuming \(k
eq 2\) and \(k
eq -1/2\) (to ensure denominators are not zero), we can cancel the common factors \((2k+1)\) and \((k-2)\) from both the numerator and the denominator. This leaves us with:
\[\frac{2}{k+2}\]
- Conclusion
By comparing the result \(\frac{2}{k+2}\) with the given options, we identify that the last option is the correct one.
The simplified expression is \(\frac{2}{k+2}\). Thus, the correct choice is the fourth option, \(\boxed{\frac{2}{k+2}}\).
Examples
Algebraic simplification is a fundamental skill used in various fields like engineering and economics. For instance, when analyzing a complex circuit, you might have a formula representing total resistance that looks messy. By simplifying it using factoring and canceling like terms, you can more easily understand how changing one component value (like variable \(k\)) affects the entire system, allowing for quicker design adjustments.