Solve the inequality by factoring. 3x2−7x+2<0 Note: Enter your answer and SHOW ALL THE STEPS NECESSARY to solve this problem in the space provided. You will not receive credit without all work shown.

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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:

  1. \(3x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{3}\),
  2. \(x - 2 = 0 \quad \Rightarrow \quad x = 2\).

The critical points are \(x = \frac{1}{3}\) and \(x = 2\).

Step 6: Test intervals

We will create intervals based on these critical points:

  1. \( (-\infty, \frac{1}{3}) \)
  2. \( \left(\frac{1}{3}, 2\right) \)
  3. \( (2, \infty) \)

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: \[ \boxed{\left(\frac{1}{3}, 2\right)}. \]