\[\frac{8a^2 + 4a + 2}{2a + 3} \div \frac{8a^3 - 1}{4a^2 + 12a + 9}\]

asked by guest
on Sep 20, 2024 at 10:33 pm



You asked:

Evaluate the expression: \(\frac{\frac{8 \cdot {a}^{2} + 4 a + 2}{2 a + 3}}{\frac{8 \cdot {a}^{3} - 1}{4 \cdot {a}^{2} + 12 a + 9}}\)

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

\[\frac{\frac{8 \cdot {a}^{2} + 4 a + 2}{2 a + 3}}{\frac{8 \cdot {a}^{3} - 1}{4 \cdot {a}^{2} + 12 a + 9}} = \frac{2 \cdot \left(2 a + 3\right)}{2 a - 1}\]

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