To factorize \(3x^3 - x^2 - 3x + 1\), we can use grouping and trial methods like synthetic division or the Rational Root Theorem. Let's go step-by-step.

### Step 1: Rational Root Theorem
Using the Rational Root Theorem, the possible rational roots are the factors of the constant term (+1) divided by the factors of the leading coefficient (3). So, the possible rational roots are:
\[
\pm 1, \pm \frac{1}{3}
\]

### Step 2: Testing Roots
Let's test \(x = 1\):
\[
f(1) = 3(1)^3 - (1)^2 - 3(1) + 1 = 3 - 1 - 3 + 1 = 0
\]
So, \(x = 1\) is a root.

### Step 3: Synthetic Division
Now that we know \(x = 1\) is a root, we can divide \(3x^3 - x^2 - 3x + 1\) by \(x - 1\) using synthetic division.

\[
\begin{array}{r|rrrr}
1 & 3 & -1 & -3 & 1 \\
  &   & 3  &  2  & -1 \\
\hline
  & 3 &  2  & -1  & 0 \\
\end{array}
\]

The quotient is \(3x^2 + 2x - 1\), so:
\[
3x^3 - x^2 - 3x + 1 = (x - 1)(3x^2 + 2x - 1)
\]

### Step 4: Factor the Quadratic
Now, we need to factor \(3x^2 + 2x - 1\). We look for two numbers that multiply to \(3 \times (-1) = -3\) and add to 2. These numbers are 3 and -1.

So, we can rewrite the quadratic as:
\[
3x^2 + 2x - 1 = 3x^2 + 3x - x - 1
\]
Now, factor by grouping:
\[
= 3x(x + 1) - 1(x + 1)
\]
\[
= (3x - 1)(x + 1)
\]

### Final Answer:
The complete factorization is:
\[
3x^3 - x^2 - 3x + 1 = (x - 1)(3x - 1)(x + 1)
\]

Would you like further details or have any questions?

### Related Questions:
1. How can you verify if a factorization is correct?
2. What other methods can be used to factor cubic equations?
3. How does synthetic division work in more complex problems?
4. What if the Rational Root Theorem yields no rational roots?
5. How can we apply factorization to solve polynomial equations?

### Tip:
Always verify factorizations by multiplying the factors back together to ensure you get the original polynomial.

Learn how to factor the cubic polynomial 3x^3 - x^2 - 3x + 1 using the Rational Root Theorem and synthetic division. This step-by-step guide explains the factorization process, making it suitable for high school students in Grades 10-12.

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