To solve the equation \( 0 = 3x^2 - 4x + 1 \) by factoring, we'll first try to factor the quadratic expression.
The given equation can be rearranged as:
\[ 3x^2 - 4x + 1 = 0 \]
To factor this, we need two numbers that multiply to \( 3 \times 1 = 3 \) (the product of the leading coefficient and the constant term) and add to \(-4\) (the coefficient of the linear term).
The numbers that meet these criteria are \(-3\) and \(-1\). Therefore, we can rewrite \(-4x\) as \(-3x - x\):
\[ 3x^2 - 3x - x + 1 = 0 \]
Now, we can group the terms:
\[ (3x^2 - 3x) + (-x + 1) = 0 \]
Factoring out common factors from each group gives:
\[ 3x(x - 1) - 1(x - 1) = 0 \]
Now we can factor by grouping:
\[ (3x - 1)(x - 1) = 0 \]
Setting each factor equal to zero gives us the solutions:
- \( 3x - 1 = 0 \) ⇒ \( 3x = 1 \) ⇒ \( x = \frac{1}{3} \)
- \( x - 1 = 0 \) ⇒ \( x = 1 \)
You provided one of the solutions as \( x = \frac{1}{3} \). Therefore, the other solution is:
\[ x = 1 \]
Thus, the other solution is \( x = 1 \).