Solution:

Given: $x^2 - 5x + 6 \leq 0$

Step 1: Factor the polynomial $x^2 - 5x + 6 = (x - 2)(x - 3)$

Step 2: Find critical values

Set each factor equal to zero:

• $x - 2 = 0 \Rightarrow x = 2$
• $x - 3 = 0 \Rightarrow x = 3$

Critical values: $x = 2, 3$

Step 3: Create a sign chart

Test intervals: $(-\infty, 2)$, $(2, 3)$, $(3, \infty)$

Step 4: Identify where $f(x) \leq 0$

We need intervals where the product is negative or zero.

From the chart:

• Negative in interval $(2, 3)$
• Zero at $x = 2$ and $x = 3$

Since we have $\leq$ (includes equality), we include the critical values.

Answer: $[2, 3]$

Verification:

• At $x = 2$: $(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$ ✓
• At $x = 2.5$: $(2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 < 0$ ✓
• At $x = 3$: $(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$ ✓

Solution:

Given: $x^3 - 4x > 0$

Step 1: Factor the polynomial $x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2)$

Step 2: Find critical values

Set each factor equal to zero:

• $x = 0$
• $x - 2 = 0 \Rightarrow x = 2$
• $x + 2 = 0 \Rightarrow x = -2$

Critical values: $x = -2, 0, 2$ (in order)

Step 3: Create a sign chart

Test intervals: $(-\infty, -2)$, $(-2, 0)$, $(0, 2)$, $(2, \infty)$

Step 4: Identify where the product is positive

We need $> 0$ (strictly positive, exclude zeros).

From the chart, the product is positive in:

• $(-2, 0)$
• $(2, \infty)$

Since we have $>$ (strict inequality), we exclude the critical values.

Answer: $(-2, 0) \cup (2, \infty)$

Verification:

• At $x = -1$ (in first interval): $(-1)^3 - 4(-1) = -1 + 4 = 3 > 0$ ✓
• At $x = 1$ (in middle): $(1)^3 - 4(1) = 1 - 4 = -3 < 0$ ✓
• At $x = 3$ (in last interval): $(3)^3 - 4(3) = 27 - 12 = 15 > 0$ ✓

Solution:

Given: $(x - 1)^2(x + 3) \geq 0$

Step 1: Identify factors (already factored)

The polynomial is already factored:

• $(x - 1)^2$ with multiplicity 2
• $(x + 3)$ with multiplicity 1

Step 2: Find critical values

• $x - 1 = 0 \Rightarrow x = 1$ (multiplicity 2)
• $x + 3 = 0 \Rightarrow x = -3$ (multiplicity 1)

Critical values: $x = -3, 1$

Step 3: Analyze signs

Important: $(x - 1)^2$ is always non-negative and equals 0 only at $x = 1$.

Since $(x - 1)^2 \geq 0$ always, the sign of the product depends on $(x + 3)$:

Step 4: Identify where product $\geq 0$

We need non-negative values (positive or zero).

From analysis:

• Negative when $x < -3$
• Zero at $x = -3$ (included with $\geq$)
• Positive when $-3 < x < 1$
• Zero at $x = 1$ (included with $\geq$)
• Positive when $x > 1$

Answer: $[-3, \infty)$

Key insight: The even multiplicity at $x = 1$ means the sign doesn't change there. The factor $(x-1)^2$ touches the x-axis but doesn't cross it.

Verification:

• At $x = -4$: $(-4-1)^2(-4+3) = 25(-1) = -25 < 0$ ✓
• At $x = -3$: $(-3-1)^2(-3+3) = 16(0) = 0$ ✓
• At $x = 0$: $(0-1)^2(0+3) = 1(3) = 3 > 0$ ✓
• At $x = 1$: $(1-1)^2(1+3) = 0(4) = 0$ ✓
• At $x = 2$: $(2-1)^2(2+3) = 1(5) = 5 > 0$ ✓