85. \( \frac{8 x^{2}+18 x-5}{10 x^{2}-9 x+2} \div \frac{8 x^{2}+22 x+15}{10 x^{2}+11 x-6} \)

To simplify the expression \( \frac{8 x^{2}+18 x-5}{10 x^{2}-9 x+2} \div \frac{8 x^{2}+22 x+15}{10 x^{2}+11 x-6} \), we start by rewriting the division as multiplication by the reciprocal: \[ \frac{8 x^{2}+18 x-5}{10 x^{2}-9 x+2} \times \frac{10 x^{2}+11 x-6}{8 x^{2}+22 x+15} \] Next, let's factor each polynomial: 1. For \( 8x^2 + 18x - 5 \), we can factor it using the method of splitting the middle term: \[ 8x^2 + 20x - 2x - 5 = 4x(2x + 5) - 1(2x + 5) = (4x - 1)(2x + 5) \] 2. For \( 10x^2 - 9x + 2 \), let's find the factors: \[ 10x^2 - 5x - 4x + 2 = 5x(2x - 1) - 2(2x - 1) = (5x - 2)(2x - 1) \] 3. Next factor \( 8x^2 + 22x + 15 \): \[ 8x^2 + 20x + 2x + 15 = 4x(2x + 5) + 3(2x + 5) = (4x + 3)(2x + 5) \] 4. Finally, factor \( 10x^2 + 11x - 6 \): \[ 10x^2 + 15x - 4x - 6 = 5x(2x + 3) - 2(2x + 3) = (5x - 2)(2x + 3) \] Substituting the factored polynomials back into our expression: \[ \frac{(4x - 1)(2x + 5)}{(5x - 2)(2x - 1)} \times \frac{(5x - 2)(2x + 3)}{(4x + 3)(2x + 5)} \] Now we can cancel the common terms: - \( (2x + 5) \) cancels with \( (2x + 5) \) - \( (5x - 2) \) cancels with \( (5x - 2) \) The expression simplifies to: \[ \frac{(4x - 1)(2x + 3)}{(4x + 3)(2x - 1)} \] So, the final simplified expression is: \[ \frac{(4x - 1)(2x + 3)}{(4x + 3)(2x - 1)} \]