To solve the inequality $|x^2 - 5x + 3| \leq 3$, we proceed as follows:

Step 1: Break the Absolute Value Inequality

The inequality $|x^2 - 5x + 3| \leq 3$ means: $-3 \leq x^2 - 5x + 3 \leq 3.$

Step 2: Solve the Compound Inequality

We solve the two parts of this inequality separately:

1. $x^2 - 5x + 3 \geq -3$,
2. $x^2 - 5x + 3 \leq 3$.

Solve $x^2 - 5x + 3 \geq -3$:

$x^2 - 5x + 3 + 3 \geq 0 \implies x^2 - 5x + 6 \geq 0.$ Factoring: $x^2 - 5x + 6 = (x - 2)(x - 3).$ The inequality becomes: $(x - 2)(x - 3) \geq 0.$ This quadratic is zero at $x = 2$ and $x = 3$. Using a sign chart, the intervals are: $x \in (-\infty, 2] \cup [3, \infty).$

Solve $x^2 - 5x + 3 \leq 3$:

$x^2 - 5x + 3 - 3 \leq 0 \implies x^2 - 5x \leq 0.$ Factoring: $x(x - 5) \leq 0.$ This quadratic is zero at $x = 0$ and $x = 5$. Using a sign chart, the intervals are: $x \in [0, 5].$

Step 3: Combine the Solutions

The overall solution is the intersection of the two conditions: $x \in (-\infty, 2] \cup [3, \infty) \quad \text{and} \quad x \in [0, 5].$ Taking the intersection:

• On $(-\infty, 2]$, intersecting with $[0, 5]$, we get $[0, 2]$.
• On $[3, \infty)$, intersecting with $[0, 5]$, we get $[3, 5]$.

Thus, the solution is: [ x \in [0, 2] \cup [3, 5]. ]

Final Answer:

[ \boxed{x \in [0, 2] \cup [3, 5]} ]

Would you like a step-by-step explanation of any part? Here are related questions to explore:

1. How do you solve absolute value inequalities in general?
2. What is the process of solving compound inequalities?
3. How do sign charts help in solving polynomial inequalities?
4. What are the properties of absolute value inequalities?
5. How does factoring help in solving quadratic inequalities?

Tip: Always check the intersection of solution sets when dealing with compound inequalities involving absolute values.