Express the following as a single algebr \( \frac{2 a+4 b}{(a+3 b)^{2}} \times \frac{a^{2}+a b-6 b^{2}}{a^{2}-4 b^{2}} \)

To express the given algebraic expression as a single fraction, we first need to simplify each part before multiplying them together: The expression is: \[ \frac{2 a + 4 b}{(a + 3 b)^{2}} \times \frac{a^{2} + ab - 6b^{2}}{a^{2} - 4b^{2}} \] Let's simplify \(2a + 4b\): \[ 2a + 4b = 2(a + 2b) \] Now, we'll look at \(a^2 + ab - 6b^2\) and \(a^2 - 4b^2\). The polynomial \(a^2 - 4b^2\) can be factored using the difference of squares: \[ a^2 - 4b^2 = (a - 2b)(a + 2b) \] Next, we will see if \(a^2 + ab - 6b^2\) can be factored. To do this, we can look for two numbers that multiply to \(-6b^2\) and add to \(b\). The numbers \(3b\) and \(-2b\) work, so we can factor it: \[ a^2 + ab - 6b^2 = (a + 3b)(a - 2b) \] Now, substituting all our simplifications back into the expression: \[ \frac{2(a + 2b)}{(a + 3b)^{2}} \times \frac{(a + 3b)(a - 2b)}{(a - 2b)(a + 2b)} \] Now, we can simplify: - The \((a + 3b)\) from the numerator of the second fraction cancels with one \((a + 3b)\) in the denominator. - The \((a - 2b)\) in the numerator cancels with one in the denominator. - The \((a + 2b)\) cancels with the numerator. Finally, the expression simplifies to: \[ \frac{2}{a + 3b} \] So, the final answer is: \[ \frac{2}{a + 3b} \]