1. The problem is to express the equation $\frac{1}{x-2} - \frac{2}{x+5} = \frac{3}{x+1}$ in the form $ax^2 + bx + c = 0$. 2. To do this, we first find a common denominator for the fractions, which is $(x-2)(x+5)(x+1)$. 3. Multiply both sides of the equation by the common denominator to clear the fractions: $$ (x+5)(x+1) - 2(x-2)(x+1) = 3(x-2)(x+5) $$ 4. Expand each term: - $(x+5)(x+1) = x^2 + 6x + 5$ - $2(x-2)(x+1) = 2(x^2 - x - 2) = 2x^2 - 2x - 4$ - $3(x-2)(x+5) = 3(x^2 + 3x - 10) = 3x^2 + 9x - 30$ 5. Substitute back into the equation: $$ x^2 + 6x + 5 - (2x^2 - 2x - 4) = 3x^2 + 9x - 30 $$ 6. Simplify the left side: $$ x^2 + 6x + 5 - 2x^2 + 2x + 4 = -x^2 + 8x + 9 $$ 7. So the equation becomes: $$ -x^2 + 8x + 9 = 3x^2 + 9x - 30 $$ 8. Bring all terms to one side: $$ -x^2 + 8x + 9 - 3x^2 - 9x + 30 = 0 $$ 9. Combine like terms: $$ -4x^2 - x + 39 = 0 $$ 10. The equation in the form $ax^2 + bx + c = 0$ is: $$ -4x^2 - x + 39 = 0 $$ This is the required quadratic form.