Question

Solve the quadratic inequality 3x2+3x−6≤0 by factoring.(1 point)
Responses

−2≤x≤1
negative 2 less-than-or-equal-to x less-than-or-equal-to 1

−1<x<2
negative 1 less than x less than 2

x≤−2 or x≥3
x less-than-or-equal-to negative 2 or x greater-than-or-equal-to 3

−3<x<1

Answer 1

To solve the quadratic inequality \( 3x^2 + 3x - 6 \leq 0 \) by factoring, let's first simplify the inequality:

1. Divide the entire inequality by 3:
\[
x^2 + x - 2 \leq 0
\]

2. 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:
\[
(x + 2)(x - 1) \leq 0
\]

3. 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 \( x = 1 \).

4. 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:
- \( (-\infty, -2) \)
- \( (-2, 1) \)
- \( (1, \infty) \)

- **For \( x < -2 \) (e.g., \( x = -3 \)):**
\[
(-3 + 2)(-3 - 1) = (-1)(-4) = 4 > 0 \quad \text{(not part of the solution set)}
\]

- **For \( -2 < x < 1 \) (e.g., \( x = 0 \)):**
\[
(0 + 2)(0 - 1) = (2)(-1) = -2 < 0 \quad \text{(part of the solution set)}
\]

- **For \( x > 1 \) (e.g., \( x = 2 \)):**
\[
(2 + 2)(2 - 1) = (4)(1) = 4 > 0 \quad \text{(not part of the solution set)}
\]

5. 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:
\[
-2 \leq x \leq 1
\]