Show that \( 2 x+1 \) is a factor of \( 6 x^{3}-5 x^{2}-12 x-4 \). Then find the roots of \( 6 x^{3}-5 x^{2}-12 x-4=0 \)

To determine whether \( 2x + 1 \) is a factor of the polynomial \( 6x^3 - 5x^2 - 12x - 4 \) and to find all roots of the equation \( 6x^3 - 5x^2 - 12x - 4 = 0 \), follow these steps: ### 1. Verify that \( 2x + 1 \) is a Factor **Factor Theorem:** A polynomial \( P(x) \) has \( (ax + b) \) as a factor if and only if \( P\left(-\frac{b}{a}\right) = 0 \). For \( 2x + 1 \): \[ 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \] Evaluate \( P\left(-\frac{1}{2}\right) \): \[ P\left(-\frac{1}{2}\right) = 6\left(-\frac{1}{2}\right)^3 - 5\left(-\frac{1}{2}\right)^2 - 12\left(-\frac{1}{2}\right) - 4 \] \[ = 6\left(-\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) + 6 - 4 \] \[ = -\frac{6}{8} - \frac{5}{4} + 2 \] \[ = -\frac{3}{4} - \frac{5}{4} + 2 = -2 + 2 = 0 \] Since \( P\left(-\frac{1}{2}\right) = 0 \), \( 2x + 1 \) is indeed a factor of \( P(x) \). ### 2. Factor the Polynomial Now, divide \( P(x) \) by \( 2x + 1 \) to find the other factors. **Polynomial Division:** \[ P(x) = (2x + 1)(6x^2 - 8x - 8) \] ### 3. Find the Remaining Roots Now, solve the quadratic equation \( 6x^2 - 8x - 8 = 0 \): First, simplify by dividing by 2: \[ 3x^2 - 4x - 4 = 0 \] Use the quadratic formula: \[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \] \[ x = \frac{4 \pm \sqrt{16 + 48}}{6} \] \[ x = \frac{4 \pm \sqrt{64}}{6} \] \[ x = \frac{4 \pm 8}{6} \] Thus, the solutions are: \[ x = \frac{12}{6} = 2 \quad \text{and} \quad x = \frac{-4}{6} = -\frac{2}{3} \] ### 4. Summary of All Roots The roots of the equation \( 6x^3 - 5x^2 - 12x - 4 = 0 \) are: \[ x = -\frac{1}{2}, \quad x = 2, \quad \text{and} \quad x = -\frac{2}{3} \] **Final Answer:** Yes, \(2x + 1\) is a factor. The equation \(6x^3 - 5x^2 - 12x - 4 = 0\) has roots \(x = -\tfrac{1}{2}\), \(x = 2\), and \(x = -\tfrac{2}{3}\).