To solve the inequality \( 3x^2 - 7x + 2 < 0 \) by factoring, we follow these steps:
Step 1: Factor the quadratic expression
We want to factor the quadratic \( 3x^2 - 7x + 2 \). We will look for two numbers that multiply to \( 3 \cdot 2 = 6 \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( -7 \) (the coefficient of \( x \)).
The numbers that meet these criteria are \( -6 \) and \( -1 \) because:
- \( -6 \cdot -1 = 6 \)
- \( -6 + (-1) = -7 \)
Now we can rewrite the quadratic expression:
\[ 3x^2 - 6x - 1x + 2 \]
Next, we group the terms:
\[ (3x^2 - 6x) + (-1x + 2) \]
Factoring each group gives:
\[ 3x(x - 2) - 1(x - 2) \]
Now, we factor out \( (x - 2) \):
\[ (3x - 1)(x - 2) \]
Step 2: Set the factored expression to zero
To find the critical points, we set each factor equal to zero:
- \( 3x - 1 = 0 \) gives \( x = \frac{1}{3} \)
- \( x - 2 = 0 \) gives \( x = 2 \)
Step 3: Determine the intervals for testing
The critical points divide the number line into three intervals:
- \( (-\infty, \frac{1}{3}) \)
- \( \left(\frac{1}{3}, 2\right) \)
- \( (2, \infty) \)
Step 4: Test points from each interval
Let's test a point from each interval to determine where the expression \( (3x - 1)(x - 2) < 0 \).
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For \( x = 0 \) (in \( (-\infty, \frac{1}{3}) \)): \[ (3(0) - 1)(0 - 2) = (-1)(-2) = 2 > 0 \] The product is positive.
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For \( x = 1 \) (in \( \left(\frac{1}{3}, 2\right) \)): \[ (3(1) - 1)(1 - 2) = (3 - 1)(1 - 2) = (2)(-1) = -2 < 0 \] The product is negative.
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For \( x = 3 \) (in \( (2, \infty) \)): \[ (3(3) - 1)(3 - 2) = (9 - 1)(3 - 2) = (8)(1) = 8 > 0 \] The product is positive.
Step 5: Combine the results
The product \( (3x - 1)(x - 2) < 0 \) is true in the interval \( \left(\frac{1}{3}, 2\right) \).
Therefore, the solution to the inequality \( 3x^2 - 7x + 2 < 0 \) is:
\[ \boxed{\left(\frac{1}{3}, 2\right)} \]