If a, b, c are real numbers such that $ \frac{3 a+2 b}{c+d}+\frac{3}{2}=0 $ then the equation $ a x^{3}+b x^{2}+c x+d=0 $ has

Define the polynomial function f(x) = ax³ + bx² + cx + d. By evaluating the polynomial at specific values, we see that f(2) = 8a + 4b + 2c + d. Evaluating f(-2) gives f(-2) = -8a + 4b - 2c + d. By manipulating the given condition algebraically, we can form the relation (3a + 2b) + (c + d)(3/2) = 0, which transforms into 6a + 4b + 3c + 3d = 0. Since the original equation relates linearly to the f(-2) expression, the Intermediate Value Theorem ensures the graph crosses the x-axis somewhere in the interval [-2, 0]. Therefore, the equation has at least one root in [-2, 0].

• Thanks
• Wrong explanation