PART 2: Solving Inequalities – VideoThis part of the project requires you to record a video (max 6 min) that covers the concept of solvinginequalities. In your video, you must go through detailed solutions of two example questions from thelists provided below and show how to solve that inequality by using one of the following methods: i)using intervals and ii) considering all cases. Show your solutions on a number line.Choose the 1stinequality from this list:1. 2x3+ 3x2 − 2x − 3 < 02. x3 − 2x2 − 5x + 6 > 03. x3 − 5x2+ 2x + 8 < 04. (x + 2)(3 − x)(x + 1) > 0Choose the 2ndinequality from this list:1.x2+9x+14x2−6x+5< 02.x−2x <x−4x−63.x2−3x−4x2+11x+30≤04.5x+4 ≥2x+1

Claro, aquí tienes una guía paso a paso para resolver las desigualdades seleccionadas utilizando los métodos mencionados y representando las soluciones en una recta numérica.

Primera Desigualdad: $x^3 - 2x^2 - 5x + 6 > 0$

Paso 1: Encontrar las raíces del polinomio

Para resolver $x^3 - 2x^2 - 5x + 6 = 0$, podemos usar el método de factorización o el teorema del factor. Probemos con valores posibles de $x$:

• $x = 1$: $1^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0$

Entonces, $x = 1$ es una raíz. Ahora, factorizamos el polinomio usando $(x - 1)$:

$x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6)$

Factorizamos $x^2 - x - 6$:

$x^2 - x - 6 = (x - 3)(x + 2)$

Entonces, la factorización completa es:

$x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2)$

Paso 2: Determinar los intervalos

Las raíces son $x = 1$, $x = -2$ y $x = 3$. Estas raíces dividen la recta numérica en los siguientes intervalos:

• $(-\infty, -2)$
• $(-2, 1)$
• $(1, 3)$
• $(3, \infty)$

Paso 3: Probar los intervalos

Probamos un valor en cada intervalo para determinar el signo del producto:

• Para $x = -3$ en $(-\infty, -2)$: $(-3 - 1)(-3 - 3)(-3 + 2) = (-4)(-6)(-1) = -24$ (negativo)

• Para $x = 0$ en $(-2, 1)$: $(0 - 1)(0 - 3)(0 + 2) = (-1)(-3)(2) = 6$ (positivo)

• Para $x = 2$ en $(1, 3)$: $(2 - 1)(2 - 3)(2 + 2) = (1)(-1)(4) = -4$ (negativo)

• Para $x = 4$ en $(3, \infty)$: $(4 - 1)(4 - 3)(4 + 2) = (3)(1)(6) = 18$ (positivo)

Paso 4: Solución en la recta numérica

La desigualdad $x^3 - 2x^2 - 5x + 6 > 0$ es positiva en los intervalos $(-2, 1)$ y $(3, \infty)$.

Segunda Desigualdad: $\frac{x^2 + 9x + 14}{x^2 - 6x + 5} < 0$

Paso 1: Factorizar numerador y denominador

Numerador: $x^2 + 9x + 14 = (x + 7)(x + 2)$

Denominador: $x^2 - 6x + 5 = (x - 1)(x - 5)$

Entonces, la desigualdad se convierte en:

$\frac{(x + 7)(x + 2)}{(x - 1)(x - 5)} < 0$

Paso 2: Encontrar las raíces y puntos críticos

Las raíces del numerador son $x = -7$ y $x = -2$. Las raíces del denominador son $x = 1$ y $x = 5$.

Sure, I can help you with that. Let's go through the steps to solve two inequalities, one from each list, using the methods specified: intervals and considering all cases. We will also represent the solutions on a number line.

First Inequality: $(x + 2)(3 - x)(x + 1) > 0$

Step-by-Step Solution:

1. Identify the critical points: These are the values of $x$ that make each factor zero.

   • $x + 2 = 0$ ⟹ $x = -2$
   • $3 - x = 0$ ⟹ $x = 3$
   • $x + 1 = 0$ ⟹ $x = -1$

2. Divide the number line into intervals based on these critical points:

   • $(-\infty, -2)$
   • $(-2, -1)$
   • $(-1, 3)$
   • $(3, \infty)$

3. Test each interval to determine where the inequality holds true. Choose a test point from each interval and substitute it into the inequality $(x + 2)(3 - x)(x + 1) > 0$.

   • For $x \in (-\infty, -2)$, choose $x = -3$: $(-3 + 2)(3 - (-3))(-3 + 1) = (-1)(6)(-2) = 12 > 0$ This interval satisfies the inequality.

   • For $x \in (-2, -1)$, choose $x = -1.5$: $(-1.5 + 2)(3 - (-1.5))(-1.5 + 1) = (0.5)(4.5)(-0.5) = -1.125 < 0$ This interval does not satisfy the inequality.

   • For $x \in (-1, 3)$, choose $x = 0$: $(0 + 2)(3 - 0)(0 + 1) = (2)(3)(1) = 6 > 0$ This interval satisfies the inequality.

   • For $x \in (3, \infty)$, choose $x = 4$: $(4 + 2)(3 - 4)(4 + 1) = (6)(-1)(5) = -30 < 0$ This interval does not satisfy the inequality.

4. Combine the intervals where the inequality holds true: $x \in (-\infty, -2) \cup (-1, 3)$

5. Represent the solution on a number line:

   • Draw a number line and mark the critical points $-2, -1,$ and $3$.
   • Shade the intervals $(-\infty, -2)$ and $(-1, 3)$.

Second Inequality: $\frac{x^2 + 9x + 14}{x^2 - 6x + 5} < 0$

Step-by-Step Solution:

1. Factorize the numerator and the denominator:

   • Numerator: $x^2 + 9x + 14 = (x + 7)(x + 2)$
   • Denominator: $x^2 - 6x + 5 = (x - 1)(x - 5)$

2. Identify the critical points: These are the values of $x$ that make the numerator or the denominator zero.

   • $x + 7 = 0$ ⟹ $x = -7$
   • $x + 2 = 0$ ⟹ $x = -2$
   • $x - 1 = 0$ ⟹ $x = 1$
   • $x - 5 = 0$ ⟹ $x = 5$

3. Divide the number line into intervals based on these critical points:

   • $(-\infty, -7)$
   • $(-7, -2)$
   • $(-2, 1)$
   • $(1, 5)$
   • $(5, \infty)$

4. Test each interval to determine where the inequality holds true. Choose a test point from each interval and substitute it into the inequality $\frac{(x + 7)(x + 2)}{(x - 1)(x - 5)} < 0$.

   • For $x \in (-\infty, -7)$, choose $x = -8$: $\frac{(-8 + 7)(-8 + 2)}{(-8 - 1)(-8 - 5)} = \frac{(-1)(-6)}{(-9)(-13)} = \frac{6}{117} > 0$ This interval does not satisfy the inequality.

   • For $x \in (-7, -2)$, choose $x = -5$: $\frac{(-5 + 7)(-5 + 2)}{(-5 - 1)(-5 - 5)} = \frac{(2)(-3)}{(-6)(-10)} = \frac{-6}{60} < 0$ This interval satisfies the inequality.

   • For $x \in (-2, 1)$, choose $x = 0$: $\frac{(0 + 7)(0 + 2)}{(0 - 1)(0 - 5)} = \frac{(7)(2)}{(-1)(-5)} = \frac{14}{5} > 0$ This interval does not satisfy the inequality.

   • For $x \in (1, 5)$, choose $x = 3$: $\frac{(3 + 7)(3 + 2)}{(3 - 1)(3 - 5)} = \frac{(10)(5)}{(2)(-2)} = \frac{50}{-4} < 0$ This interval satisfies the inequality.

   • For $x \in (5, \infty)$, choose $x = 6$: $\frac{(6 + 7)(6 + 2)}{(6 - 1)(6 - 5)} = \frac{(13)(8)}{(5)(1)} = \frac{104}{5} > 0$ This interval does not satisfy the inequality.

5. Combine the intervals where the inequality holds true: $x \in (-7, -2) \cup (1, 5)$

6. Represent the solution on a number line:

   • Draw a number line and mark the critical points $-7, -2, 1,$ and $5$.
   • Shade the intervals $(-7, -2)$ and $(1, 5)$.

Summary

• For the first inequality $(x + 2)(3 - x)(x + 1) > 0$, the solution is $x \in (-\infty, -2) \cup (-1, 3)$.
• For the second inequality $\frac{x^2 + 9x + 14}{x^2 - 6x + 5} < 0$, the solution is $x \in (-7, -2) \cup (1, 5)$.

Make sure to clearly explain each step in your video and show the number line representations for both inequalities. Good luck with your project!