Question

Solve: $(x ^ 2 - 3x + 2)(x ^ 3 - 3x ^ 2)(4 - x ^ 2) \ge 0$

Answer 1

$(-\infty, -2] \cup \{0\} \cup [1, 3]$

Answer 2

The given inequality is $(x^2 - 3x + 2)(x^3 - 3x^2)(4 - x^2) \ge 0$.

First, factorize each term in the inequality:

1. $x^2 - 3x + 2 = (x-1)(x-2)$
2. $x^3 - 3x^2 = x^2(x-3)$
3. $4 - x^2 = (2-x)(2+x) = -(x-2)(x+2)$

Substitute the factored forms back into the inequality: $(x-1)(x-2) \cdot x^2(x-3) \cdot -(x-2)(x+2) \ge 0$

Rearrange the terms and combine the $(x-2)$ factors: $- x^2 (x-1)(x-2)^2 (x-3)(x+2) \ge 0$

To make the leading coefficient positive, multiply the entire inequality by $-1$ and reverse the inequality sign: $x^2 (x-1)(x-2)^2 (x-3)(x+2) \le 0$

Now, find the roots of the polynomial $Q(x) = x^2 (x-1)(x-2)^2 (x-3)(x+2)$. The roots are the values of $x$ where $Q(x) = 0$. The roots are:

• $x=0$ (from $x^2$, multiplicity 2)
• $x=1$ (from $x-1$, multiplicity 1)
• $x=2$ (from $(x-2)^2$, multiplicity 2)
• $x=3$ (from $x-3$, multiplicity 1)
• $x=-2$ (from $x+2$, multiplicity 1)

List the roots in ascending order: $-2, 0, 1, 2, 3$. These roots divide the number line into intervals: $(-\infty, -2), (-2, 0), (0, 1), (1, 2), (2, 3), (3, \infty)$.

We need to find the intervals where $Q(x) = x^2 (x-1)(x-2)^2 (x-3)(x+2) \le 0$. We can use the sign chart method or the wavy curve method. The leading coefficient of $Q(x)$ is positive (it's 1). So, for $x > 3$, $Q(x) > 0$.

Now, consider the sign changes at each root, moving from right to left:

• At $x=3$ (multiplicity 1, odd): The sign changes from positive to negative. So, $Q(x) < 0$ in $(2, 3)$.
• At $x=2$ (multiplicity 2, even): The sign does not change. So, $Q(x) < 0$ in $(1, 2)$.
• At $x=1$ (multiplicity 1, odd): The sign changes from negative to positive. So, $Q(x) > 0$ in $(0, 1)$.
• At $x=0$ (multiplicity 2, even): The sign does not change. So, $Q(x) > 0$ in $(-2, 0)$.
• At $x=-2$ (multiplicity 1, odd): The sign changes from positive to negative. So, $Q(x) < 0$ in $(-\infty, -2)$.

Summary of the sign of $Q(x)$:

• $(-\infty, -2)$: $Q(x) < 0$
• $(-2, 0)$: $Q(x) > 0$
• $(0, 1)$: $Q(x) > 0$
• $(1, 2)$: $Q(x) < 0$
• $(2, 3)$: $Q(x) < 0$
• $(3, \infty)$: $Q(x) > 0$

We are looking for $Q(x) \le 0$. This means the intervals where $Q(x) < 0$ and the roots where $Q(x) = 0$. The intervals where $Q(x) < 0$ are $(-\infty, -2)$, $(1, 2)$, and $(2, 3)$. This can be written as $(-\infty, -2) \cup (1, 3)$ excluding $x=2$. The roots where $Q(x) = 0$ are $\{-2, 0, 1, 2, 3\}$.

Combining the intervals where $Q(x) < 0$ and the roots where $Q(x) = 0$:

• $(-\infty, -2) \cup \{-2\} = (-\infty, -2]$
• $(1, 2) \cup \{1, 2\} = [1, 2]$
• $(2, 3) \cup \{2, 3\} = [2, 3]$

The union of $[1, 2]$ and $[2, 3]$ is $[1, 3]$. So, the solution set from the intervals $(-\infty, -2)$, $(1, 2)$, $(2, 3)$ and the roots $\{-2, 1, 2, 3\}$ is $(-\infty, -2] \cup [1, 3]$.

We also need to include the root $x=0$, which makes $Q(0)=0$ and satisfies $Q(x) \le 0$. The root $x=0$ is not included in $(-\infty, -2]$ or $[1, 3]$.

Therefore, the complete solution set is $(-\infty, -2] \cup \{0\} \cup [1, 3]$.