To solve the inequality \(3x^2 - 7x + 2 < 0\) by factoring, follow these steps:
Step 1: Factor the quadratic expression
We start with the expression \(3x^2 - 7x + 2\). We need to factor this quadratic.
We look for two numbers that multiply to \(3 \times 2 = 6\) (the product of the leading coefficient and the constant term) and add to \(-7\) (the coefficient of the linear term).
The two numbers that fulfill these conditions are \(-1\) and \(-6\) because: \[ -1 \times -6 = 6 \quad \text{and} \quad -1 + (-6) = -7. \]
Step 2: Rewrite the middle term
We can rewrite \(-7x\) using \(-1x\) and \(-6x\): \[ 3x^2 - x - 6x + 2. \]
Step 3: Factor by grouping
Now we will group the terms: \[ (3x^2 - x) + (-6x + 2). \]
Next, factor out the common factors in each group: \[ x(3x - 1) - 2(3x - 1). \]
Now factor out the common binomial \((3x - 1)\): \[ (3x - 1)(x - 2). \]
Step 4: Set up the inequality
Now we rewrite our inequality using the factored form: \[ (3x - 1)(x - 2) < 0. \]
Step 5: Find the critical points
We set each factor to zero to find the critical points:
- \(3x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{3}\),
- \(x - 2 = 0 \quad \Rightarrow \quad x = 2\).
The critical points are \(x = \frac{1}{3}\) and \(x = 2\).
Step 6: Test intervals
We will create intervals based on these critical points:
- \( (-\infty, \frac{1}{3}) \)
- \( \left(\frac{1}{3}, 2\right) \)
- \( (2, \infty) \)
Now we will test each interval to see where the product \((3x - 1)(x - 2)\) is negative.
-
Interval 1: \( (-\infty, \frac{1}{3}) \)
Choose \(x = 0\): \[ (3(0) - 1)(0 - 2) = (-1)(-2) = 2 \quad \text{(positive)}. \] -
Interval 2: \( \left(\frac{1}{3}, 2\right) \)
Choose \(x = 1\): \[ (3(1) - 1)(1 - 2) = (3 - 1)(1 - 2) = (2)(-1) = -2 \quad \text{(negative)}. \] -
Interval 3: \( (2, \infty) \)
Choose \(x = 3\): \[ (3(3) - 1)(3 - 2) = (9 - 1)(3 - 2) = (8)(1) = 8 \quad \text{(positive)}. \]
Step 7: Collect results
The product \((3x - 1)(x - 2)\) is negative in the interval: \[ \left(\frac{1}{3}, 2\right). \]
Step 8: Conclusion
Since the inequality is strict (<0), we do not include the endpoints.
Thus, the solution to the inequality \(3x^2 - 7x + 2 < 0\) is: \[ \boxed{\left(\frac{1}{3}, 2\right)}. \]