86. \( \frac{10+7 x-12 x^{2}}{8 x^{2}-2 x-15} \div \frac{6 x^{2}-13 x+5}{10 x^{2}-13 x+4} \)

To simplify the expression \( \frac{10+7 x-12 x^{2}}{8 x^{2}-2 x-15} \div \frac{6 x^{2}-13 x+5}{10 x^{2}-13 x+4} \), we first rewrite the division as multiplication by the reciprocal: \[ \frac{10+7 x-12 x^{2}}{8 x^{2}-2 x-15} \times \frac{10 x^{2}-13 x+4}{6 x^{2}-13 x+5} \] Now, let's factor each polynomial. 1. **Factoring \(10 + 7x - 12x^2\):** - Rearranging gives us \(-12x^2 + 7x + 10\). - To factor this, we need two numbers that multiply to \( -120 \) (the product of \(-12\) and \(10\)) and add to \(7\). These numbers are \(15\) and \(-8\). - Thus, we can rewrite it as \(-12x^2 + 15x - 8x + 10 = -3x(4x - 5) + 2(4x - 5) = (4x - 5)(-3x + 2)\). 2. **Factoring \(8x^2 - 2x - 15\):** - The product is \((-120)\) and we want the sum to be \(-2\). The factor pair that works is \(10\) and \(-12\). - Rewriting gives \(8x^2 + 10x - 12x - 15 = 2x(4x + 5) - 3(4x + 5) = (4x + 5)(2x - 3)\). 3. **Factoring \(6x^2 - 13x + 5\):** - The product is \(30\) and we want the sum to be \(-13\). The pair is \(-10\) and \(-3\). - Rewriting gives \(6x^2 - 10x - 3x + 5 = 2x(3x - 5) - 1(3x - 5) = (3x - 5)(2x - 1)\). 4. **Factoring \(10x^2 - 13x + 4\):** - The product is \(40\) and we need \(-13\). The correct pair is \(-8\) and \(-5\). - Rearranging gives \(10x^2 - 8x - 5x + 4 = 2x(5x - 4) - 1(5x - 4) = (5x - 4)(2x - 1)\). Putting all the factors together, the expression can now be simplified: \[ \frac{(4x - 5)(-3x + 2)}{(4x + 5)(2x - 3)} \times \frac{(5x - 4)(2x - 1)}{(3x - 5)(2x - 1)} \] Now, we can cancel out like terms, specifically \((2x - 1)\) in the numerator and denominator: \[ \frac{(4x - 5)(-3x + 2)(5x - 4)}{(4x + 5)(2x - 3)(3x - 5)} \] Now, we can express the answer clearly with factored polynomials or leave it in this form, depending on the problem requirement. Thus, the final simplified result is: \[ \frac{(4x - 5)(-3x + 2)(5x - 4)}{(4x + 5)(2x - 3)(3x - 5)} \]