Express the following as a single algebraic fraction in its simplest form: \( \frac{2 a+4 b}{(a+3 b)^{2}} \times \frac{a^{2}+a b-6 b^{2}}{a^{2}-4 b^{2}} \)

To combine the fractions and simplify, we start with the expression: \[ \frac{2a + 4b}{(a + 3b)^{2}} \times \frac{a^{2} + ab - 6b^{2}}{a^{2} - 4b^{2}} \] 1. **Factor Each Part:** - The numerator \(2a + 4b\) can be factored as \(2(a + 2b)\). - The denominator \((a + 3b)^{2}\) remains unchanged. - The numerator \(a^{2} + ab - 6b^{2}\) can be factored. We look for two numbers that multiply to \(-6b^{2}\) and add to \(b\). These numbers are \(3b\) and \(-2b\), so: \[ a^{2} + ab - 6b^{2} = (a + 3b)(a - 2b) \] - The denominator \(a^{2} - 4b^{2}\) is a difference of squares: \[ a^{2} - 4b^{2} = (a - 2b)(a + 2b) \] 2. **Rewriting the Entire Expression:** This gives us: \[ \frac{2(a + 2b)}{(a + 3b)^{2}} \times \frac{(a + 3b)(a - 2b)}{(a - 2b)(a + 2b)} \] 3. **Canceling Common Factors:** Now we rewrite it as: \[ \frac{2(a + 2b)(a + 3b)(a - 2b)}{(a + 3b)^{2}(a - 2b)(a + 2b)} \] We observe that the \(a + 2b\) and \(a - 2b\) terms cancel: \[ = \frac{2(a + 3b)}{(a + 3b)(a + 3b)} \] Further simplifying gives: \[ = \frac{2}{a + 3b} \] Thus, the final simplified form of the expression is: \[ \frac{2}{a + 3b} \]