(3) Resuelua la avación polinominal, epresar la respuesta en forma gratica, en forma de intervalo y en forma de Jesiaguadad. \( -2+x^{2} \geq 3(x+0)-2 \)

Okay, I need to solve the polynomial inequality: \[ -2 + x^{2} \geq 3(x + 0) - 2 \] First, let's simplify the right side of the inequality. Since \(3(x + 0)\) simplifies to \(3x\), the inequality becomes: \[ -2 + x^{2} \geq 3x - 2 \] Next, I'll move all terms to one side to set the inequality to zero: \[ x^{2} - 3x - 2 + 2 \geq 0 \] Simplifying further: \[ x^{2} - 3x \geq 0 \] Now, I'll factor the quadratic expression: \[ x(x - 3) \geq 0 \] To find the critical points, I'll set each factor equal to zero: \[ x = 0 \quad \text{and} \quad x = 3 \] These points divide the number line into three intervals: \( (-\infty, 0) \), \( (0, 3) \), and \( (3, \infty) \). I'll test a value from each interval to determine where the inequality holds true. 1. **Interval \( (-\infty, 0) \):** Let's choose \( x = -1 \) \[ (-1)(-1 - 3) = (-1)(-4) = 4 \geq 0 \quad \text{(True)} \] 2. **Interval \( (0, 3) \):** Let's choose \( x = 1 \) \[ (1)(1 - 3) = (1)(-2) = -2 \geq 0 \quad \text{(False)} \] 3. **Interval \( (3, \infty) \):** Let's choose \( x = 4 \) \[ (4)(4 - 3) = (4)(1) = 4 \geq 0 \quad \text{(True)} \] From the tests, the inequality holds true in the intervals \( (-\infty, 0) \) and \( (3, \infty) \). Since the inequality is non-strict (\(\geq\)), the endpoints \( x = 0 \) and \( x = 3 \) are also included. **Graphical Representation:** - The graph of \( y = x^{2} - 3x \) is a parabola opening upwards with its vertex at \( x = \frac{3}{2} \). - The parabola intersects the x-axis at \( x = 0 \) and \( x = 3 \). - The inequality \( x^{2} - 3x \geq 0 \) is satisfied where the parabola is above or on the x-axis. **Interval Notation:** \[ (-\infty, 0] \cup [3, \infty) \] **Jesiguadad (Solution Set):** \[ x \leq 0 \quad \text{or} \quad x \geq 3 \]