81. \( \frac{2 x^{2}-3 x-20}{2 x^{2}-7 x-30} \div \frac{2 x^{2}-5 x-12}{4 x^{2}+12 x+9} \)

To simplify the given expression \( \frac{2 x^{2}-3 x-20}{2 x^{2}-7 x-30} \div \frac{2 x^{2}-5 x-12}{4 x^{2}+12 x+9} \), we need to rewrite the division as multiplication by the reciprocal. The expression can be rewritten as: \[ \frac{2 x^{2}-3 x-20}{2 x^{2}-7 x-30} \times \frac{4 x^{2}+12 x+9}{2 x^{2}-5 x-12} \] Next, we will factor each polynomial: 1. **For \(2 x^{2}-3 x-20\)**: To factor this, we look for two numbers that multiply to \(2 \times -20 = -40\) and add to \(-3\). These numbers are \(5\) and \(-8\). \[ 2 x^{2}-3 x-20 = 2 x^{2} + 5 x - 8 x - 20 = (2 x + 5)(x - 4) \] 2. **For \(2 x^{2}-7 x-30\)**: We want two numbers that multiply to \(2 \times -30 = -60\) and add to \(-7\): those are \(-12\) and \(5\). \[ 2 x^{2}-7 x-30 = 2 x^{2} - 12 x + 5 x - 30 = (2 x + 5)(x - 6) \] 3. **For \(2 x^{2}-5 x-12\)**: We look for two numbers that multiply to \(2 \times -12 = -24\) and add to \(-5\): these are \(-8\) and \(3\). \[ 2 x^{2}-5 x-12 = 2 x^{2} - 8 x + 3 x - 12 = (2 x + 3)(x - 4) \] 4. **For \(4 x^{2}+12 x+9\)**: This expression can be factored as a perfect square. \[ 4 x^{2}+12 x+9 = (2 x + 3)^{2} \] Now, substituting back into our expression, we have: \[ \frac{(2x + 5)(x - 4)}{(2x + 5)(x - 6)} \times \frac{(2x + 3)^{2}}{(2x + 3)(x - 4)} \] Next, we can cancel out common factors: - Cancel \(2x + 5\) from the numerator and denominator. - Cancel \(x - 4\) from the numerator and denominator. - Cancel one \(2x + 3\) from the numerator and denominator. This gives: \[ \frac{(2x + 3)}{(x - 6)} \] Thus, the simplified expression is: \[ \frac{2x + 3}{x - 6} \]