Question

matk311_mathematics iii semester a_2026-080220207-integ math 3a
multiplying and dividing rational expressions
what is the product?
\\(\frac{4k + 2}{k^2 - 4} cdot \frac{k - 2}{2k + 1}\\)
options:
\\(\frac{2}{2k + 1}\\)
\\(\frac{4}{2k + 1}\\)
\\(\frac{2}{k + 2}\\)
\\(\frac{2}{k - 2}\\)

Explanation:

Step1: Factor numerators/denominators

Factor \(4k + 2 = 2(2k + 1)\) and \(k^2 - 4=(k - 2)(k + 2)\) (difference of squares). The expression becomes \(\frac{2(2k + 1)}{(k - 2)(k + 2)}\cdot\frac{k - 2}{2k + 1}\).

Step2: Cancel common factors

Cancel \(2k + 1\) (non - zero, so valid) and \(k - 2\) (non - zero, so valid). We get \(\frac{2}{k + 2}\)? Wait, no, wait: Wait, original first fraction numerator: \(4k + 2 = 2(2k+1)\), denominator \(k^2 - 4=(k - 2)(k + 2)\). Second fraction: numerator \(k - 2\), denominator \(2k + 1\). So multiplying: \(\frac{2(2k + 1)(k - 2)}{(k - 2)(k + 2)(2k + 1)}\). Cancel \(2k + 1\) and \(k - 2\), left with \(\frac{2}{k + 2}\)? Wait, no, the first option is \(\frac{2}{2k + 1}\)? Wait, no, let's re - check. Wait, maybe I factored wrong. Wait, \(4k+2 = 2(2k + 1)\), \(k^2-4=(k - 2)(k + 2)\). Second fraction: \(\frac{k - 2}{2k + 1}\). So multiplying: \(\frac{2(2k + 1)}{(k - 2)(k + 2)}\times\frac{k - 2}{2k + 1}\). Cancel \(2k + 1\) and \(k - 2\), so we have \(\frac{2}{k + 2}\)? But the first option is \(\frac{2}{2k + 1}\)? Wait, no, maybe the problem is \(\frac{4k + 2}{k^2-4}\cdot\frac{k - 2}{2k + 1}\). Wait, \(4k + 2 = 2(2k + 1)\), \(k^2-4=(k - 2)(k + 2)\). So:

\(\frac{2(2k + 1)}{(k - 2)(k + 2)}\times\frac{k - 2}{2k + 1}=\frac{2(2k + 1)(k - 2)}{(k - 2)(k + 2)(2k + 1)}=\frac{2}{k + 2}\)? But the first option is \(\frac{2}{2k + 1}\)? Wait, no, maybe the original problem is \(\frac{4k + 2}{k - 4}\cdot\frac{k - 2}{2k + 1}\)? Wait, the user wrote \(k^2 - 4\) or \(k - 4\)? The image shows \(k^2-4\)? Wait, the user's text: \(\frac{4k + 2}{k^2-4}\cdot\frac{k - 2}{2k + 1}\). So after canceling, we get \(\frac{2}{k + 2}\), but the first option is \(\frac{2}{2k + 1}\)? Wait, no, maybe I made a mistake. Wait, let's check the options. Wait, the first option is \(\frac{2}{2k + 1}\), second \(\frac{4}{2k + 1}\), third \(\frac{2}{k + 2}\), fourth \(\frac{2}{k - 2}\). Ah! So the third option is \(\frac{2}{k + 2}\), which is the result. Wait, let's re - do the calculation:

Given \(\frac{4k + 2}{k^2-4}\cdot\frac{k - 2}{2k + 1}\)

1. Factor numerators and denominators:

• \(4k + 2=2(2k + 1)\)
• \(k^2-4=(k - 2)(k + 2)\) (by difference of squares formula \(a^2 - b^2=(a - b)(a + b)\), here \(a = k\), \(b = 2\))
• The second fraction is \(\frac{k - 2}{2k + 1}\)

1. Multiply the fractions:

• \(\frac{4k + 2}{k^2-4}\cdot\frac{k - 2}{2k + 1}=\frac{2(2k + 1)}{(k - 2)(k + 2)}\cdot\frac{k - 2}{2k + 1}\)

1. Cancel out common factors:

• We can cancel out \(2k + 1\) (assuming \(2k+1

eq0\), i.e., \(k
eq-\frac{1}{2}\)) and \(k - 2\) (assuming \(k - 2
eq0\), i.e., \(k
eq2\)) from the numerator and the denominator.

• After canceling, we are left with \(\frac{2}{k + 2}\)

Wait, but the first option in the image (the top one) is \(\frac{2}{2k + 1}\), the second is \(\frac{4}{2k + 1}\), the third is \(\frac{2}{k + 2}\), the fourth is \(\frac{2}{k - 2}\). So the correct answer is the third option \(\frac{2}{k + 2}\) (the third rectangle). But wait, maybe I misread the problem. Wait, the problem is \(\frac{4k + 2}{k^2-4}\cdot\frac{k - 2}{2k + 1}\), yes. So the product is \(\frac{2}{k + 2}\), which is the third option (the third box, labeled \(\frac{2}{k + 2}\)).

Answer:

The third option (the rectangle with \(\frac{2}{k + 2}\))