To solve the quadratic inequality \( 3x^2 + 3x - 6 \leq 0 \) by factoring, let's first simplify the inequality:
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Divide the entire inequality by 3: \[ x^2 + x - 2 \leq 0 \]
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We now need to factor \( x^2 + x - 2 \). We are looking for two numbers that multiply to -2 (the constant term) and add up to 1 (the coefficient of the linear term). The numbers that satisfy these conditions are 2 and -1. Thus, we can factor it as: \[ (x + 2)(x - 1) \leq 0 \]
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Next, we find the critical points by setting the factored expression equal to zero: \[ (x + 2)(x - 1) = 0 \] This gives us the critical points \( x = -2 \) and \( x = 1 \).
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Now we will test intervals around the critical points to see where the product \( (x + 2)(x - 1) \) is less than or equal to zero. The intervals to test are:
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\( (-\infty, -2) \)
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\( (-2, 1) \)
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\( (1, \infty) \)
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For \( x < -2 \) (e.g., \( x = -3 \)): \[ (-3 + 2)(-3 - 1) = (-1)(-4) = 4 > 0 \quad \text{(not part of the solution set)} \]
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For \( -2 < x < 1 \) (e.g., \( x = 0 \)): \[ (0 + 2)(0 - 1) = (2)(-1) = -2 < 0 \quad \text{(part of the solution set)} \]
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For \( x > 1 \) (e.g., \( x = 2 \)): \[ (2 + 2)(2 - 1) = (4)(1) = 4 > 0 \quad \text{(not part of the solution set)} \]
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Now we include the critical points because the inequality is less than or equal to zero (\( \leq \)): \[ x = -2 \] and \[ x = 1 \]
Thus, the solution to the inequality \( 3x^2 + 3x - 6 \leq 0 \) is: \[ -2 \leq x \leq 1 \]
So the correct response is: \[ -2 \leq x \leq 1 \]