83. \( \frac{9 x^{2}-16}{6 x^{2}-11 x+4} \div \frac{6 x^{2}+11 x+4}{8 x^{2}+10 x+3} \)

To simplify the expression \( \frac{9 x^{2}-16}{6 x^{2}-11 x+4} \div \frac{6 x^{2}+11 x+4}{8 x^{2}+10 x+3} \), we first rewrite the division as multiplication by the reciprocal: \[ \frac{9 x^{2}-16}{6 x^{2}-11 x+4} \times \frac{8 x^{2}+10 x+3}{6 x^{2}+11 x+4} \] Next, we factor the expressions where possible: 1. The numerator \( 9x^{2} - 16 \) can be factored using the difference of squares: \[ 9x^{2} - 16 = (3x - 4)(3x + 4) \] 2. The denominator \( 6x^{2} - 11x + 4 \) can be factored as well. We need two numbers that multiply to \( 6 \cdot 4 = 24 \) and add to -11. The numbers -3 and -8 will work, so we can factor as follows: \[ 6x^{2} - 11x + 4 = (2x - 1)(3x - 4) \] 3. For \( 6x^{2} + 11x + 4 \), we need to find two numbers that multiply to \( 6 \cdot 4 = 24 \) and add to +11, which are +3 and +8: \[ 6x^{2} + 11x + 4 = (3x + 4)(2x + 1) \] 4. Finally, factor the last term \( 8x^{2} + 10x + 3 \). Here, we find two numbers that multiply to \( 8 \cdot 3 = 24 \) and add to +10, which will be +6 and +4: \[ 8x^{2} + 10x + 3 = (4x + 3)(2x + 1) \] Now substitute the factored forms back into the expression: \[ \frac{(3x - 4)(3x + 4)}{(2x - 1)(3x - 4)} \times \frac{(4x + 3)(2x + 1)}{(3x + 4)(2x + 1)} \] Now we can cancel common factors. We note: - \( (3x - 4) \) cancels from the numerator and the denominator. - \( (3x + 4) \) cancels from the numerator and the denominator. - \( (2x + 1) \) cancels from the numerator and the denominator. After canceling, we are left with: \[ \frac{(4x + 3)}{(2x - 1)} \] Thus, the simplified expression is: \[ \frac{4x + 3}{2x - 1} \]