$\frac{8a^2+4a+2 \}{2a+3\}$ รท
$\frac{8a^3 - 1\}{4a^2 + 12a + 9\}$
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MathBot Answer:
Evaluated
\(\displaystyle \frac{\frac{8 \cdot {a}^{2} + 4 a + 2}{2 a + 3}}{\frac{8 \cdot {a}^{3} - 1}{4 \cdot {a}^{2} + 12 a + 9}} = \frac{\left(4 a^{2} + 12 a + 9\right) \left(8 a^{2} + 4 a + 2\right)}{\left(2 a + 3\right) \left(8 a^{3} - 1\right)} \)
Expanded
\[\frac{\frac{8 \cdot {a}^{2} + 4 a + 2}{2 a + 3}}{\frac{8 \cdot {a}^{3} - 1}{4 \cdot {a}^{2} + 12 a + 9}} = \frac{8 a^{2}}{\frac{16 a^{4}}{4 a^{2} + 12 a + 9} + \frac{24 a^{3}}{4 a^{2} + 12 a + 9} - \frac{2 a}{4 a^{2} + 12 a + 9} - \frac{3}{4 a^{2} + 12 a + 9}} + \frac{4 a}{\frac{16 a^{4}}{4 a^{2} + 12 a + 9} + \frac{24 a^{3}}{4 a^{2} + 12 a + 9} - \frac{2 a}{4 a^{2} + 12 a + 9} - \frac{3}{4 a^{2} + 12 a + 9}} + \frac{2}{\frac{16 a^{4}}{4 a^{2} + 12 a + 9} + \frac{24 a^{3}}{4 a^{2} + 12 a + 9} - \frac{2 a}{4 a^{2} + 12 a + 9} - \frac{3}{4 a^{2} + 12 a + 9}}\]
Factored
\[\frac{\frac{8 \cdot {a}^{2} + 4 a + 2}{2 a + 3}}{\frac{8 \cdot {a}^{3} - 1}{4 \cdot {a}^{2} + 12 a + 9}} = \frac{2 \cdot \left(2 a + 3\right)}{2 a - 1}\]