Question

multiplying/dividing rational functions
\frac{6a}{3a + 15} \cdot \frac{4a + 20}{2a^2}
\frac{40x^2}{x + 6} \div \frac{24x^6}{3x + 18}

Explanation:

Step1: Simplify the first rational function multiplication

First, factor the numerators and denominators. For \(\frac{6a}{3a + 15}\cdot\frac{4a + 20}{2a^{2}}\), factor \(3a + 15 = 3(a + 5)\) and \(4a + 20 = 4(a + 5)\). So the expression becomes \(\frac{6a}{3(a + 5)}\cdot\frac{4(a + 5)}{2a^{2}}\). Then, cancel out common factors. The 3 in the denominator and 6 in the numerator: \(6\div3 = 2\). The \(a\) in the numerator and \(a^{2}\) in the denominator: \(a\div a^{2}=\frac{1}{a}\). The \((a + 5)\) terms cancel out. The 4 and 2: \(4\div2 = 2\). So after canceling, we have \(\frac{2\times2}{a}=\frac{4}{a}\).

Step2: Simplify the second rational function division

Recall that dividing by a fraction is multiplying by its reciprocal. So \(\frac{40x^{2}}{x + 6}\div\frac{24x^{6}}{3x + 18}=\frac{40x^{2}}{x + 6}\cdot\frac{3x + 18}{24x^{6}}\). Factor \(3x + 18 = 3(x + 6)\). Now the expression is \(\frac{40x^{2}}{x + 6}\cdot\frac{3(x + 6)}{24x^{6}}\). Cancel common factors. The \((x + 6)\) terms cancel out. The 40 and 24: find the greatest common divisor (GCD) of 40 and 24, which is 8. \(40\div8 = 5\), \(24\div8 = 3\). The 3 and the remaining numbers: \(3\) in the numerator and after dividing 40 and 24, we have \(\frac{5x^{2}\times3}{3x^{6}}\) (wait, no, let's redo the factoring and canceling. \(40x^{2}\times3(x + 6)\) over \((x + 6)\times24x^{6}\). Cancel \((x + 6)\). Then, \(40\times3 = 120\), \(24\) remains. So \(\frac{120x^{2}}{24x^{6}}\). Simplify the coefficients: \(120\div24 = 5\). Simplify the variables: \(x^{2}\div x^{6}=\frac{1}{x^{4}}\). So the result is \(\frac{5}{x^{4}}\).

Answer:

For \(\boldsymbol{\frac{6a}{3a + 15}\cdot\frac{4a + 20}{2a^{2}}}\), the answer is \(\boldsymbol{\frac{4}{a}}\).
For \(\boldsymbol{\frac{40x^{2}}{x + 6}\div\frac{24x^{6}}{3x + 18}}\), the answer is \(\boldsymbol{\frac{5}{x^{4}}}\).