(2-3х)(х^2+2х-3)/3х^2+4х-4≥0

To solve the inequality \(\frac{(2-3x)(x^2+2x-3)}{3x^2+4x-4} \geq 0\), we need to follow several steps: factor the numerator and denominator, find the critical points, and determine the sign of the expression in each interval. 1. **Factor the numerator and the denominator.** - **Numerator:** \[ (2 - 3x)(x^2 + 2x - 3) \] The quadratic \(x^2 + 2x - 3\) can be factored. We need its roots: \[ x^2 + 2x - 3 = (x + 3)(x - 1) \] So, the numerator becomes: \[ (2 - 3x)(x + 3)(x - 1) \] - **Denominator:** \[ 3x^2 + 4x - 4 \] We can use the quadratic formula to find the roots: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{-4 \pm \sqrt{16 + 48}}{6} = \frac{-4 \pm \sqrt{64}}{6} = \frac{-4 \pm 8}{6} \] This provides two roots: \[ x = \frac{4}{6} = \frac{2}{3}, \quad x = \frac{-12}{6} = -2 \] Thus, the denominator factors as: \[ 3(x + 2)(x - \frac{2}{3}) \] 2. **Identify critical points.** The critical points come from the roots of the numerator and denominator: - From \((2 - 3x) = 0\): \(x = \frac{2}{3}\) - From \((x + 3) = 0\): \(x = -3\) - From \((x - 1) = 0\): \(x = 1\) - From \((3(x + 2)(x - \frac{2}{3})) = 0\): \(x = -2\) and \(x = \frac{2}{3}\) Thus, the critical points are: \[ x = -3, \quad x = -2, \quad x = \frac{2}{3}, \quad x = 1 \] 3. **Test intervals.** The critical points divide the number line into several intervals: \((- \infty, -3)\), \((-3, -2)\), \((-2, \frac{2}{3})\), \((\frac{2}{3}, 1)\), and \((1, \infty)\). We will test a value from each interval in the expression \(\frac{(2 - 3x)(x + 3)(x - 1)}{3(x + 2)(x - \frac{2}{3})}\). - **Interval \((- \infty, -3)\):** Test \(x = -4\) \[ \frac{(2 - 3(-4))((-4 + 3)(-4 - 1))}{3(-4 + 2)(-4 - \frac{2}{3})} = \frac{(14)(-1)(-5)}{3(-2)(-\frac{14}{3})} = \text{positive} \] - **Interval \((-3, -2)\):** Test \(x = -2.5\) \[ (2 - 3(-2.5))(0.5)(-3.5)/(3 \cdot (-0.5)(-2.5 - \frac{2}{3}) = \text{negative} \] - **Interval \((-2, \frac{2}{3})\):** Test \(x = 0\) \[ (2 - 0)(3)(-1)/(3(2)(-\frac{2}{3}) = \text{positive} \] - **Interval \((\frac{2}{3}, 1)\):** Test \(x = 0.8\) \[ (2 - 2.4)(3.6)(-0.2)/(3(2.8)(0.1333) = \text{negative} \] - **Interval \((1, \infty)\):** Test \(x = 2\) \[ (2 - 6)(5)(1)/(3)(4)(1.33) = \text{negative} \] 4. **Combine results and find the solution set.** So the sign of the expression is as follows: - Positive on \((- \infty, -3)\) - Negative on \((-3, -2)\) - Positive on \((-2, \frac{2}{3})\) - Negative on \((\frac{2}{3}, 1)\) - Negative on \((1, \infty)\) We now include the points where the expression is equal to zero: - The expression is undefined at \(x = -2\) and \(x = \frac{2}{3}\). - The numerator is zero at \(x = -3\), \(x = 1\). Thus, the solution to the inequality \(\frac{(2 - 3x)(x^2 + 2x - 3)}{3x^2 + 4x - 4} \geq 0\) is: \[ (-\infty, -3] \cup (-2, \frac{2}{3}) \]