Question

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

−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
negative 3 less than x less than 1

−2≤x≤1

Answer 1

To solve the quadratic inequality \( 3x^2 + 3x - 6 \leq 0 \) by factoring, we can start by simplifying the inequality.

First, divide the entire inequality by 3:

\[
x^2 + x - 2 \leq 0
\]

Next, we will factor the quadratic expression \( x^2 + x - 2 \). We need to find two numbers that multiply to \(-2\) (the constant term) and add to \(1\) (the coefficient of \(x\)). The numbers \(2\) and \(-1\) work:

\[
x^2 + x - 2 = (x + 2)(x - 1)
\]

Thus, we can rewrite the inequality as:

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

Next, we will find the critical points by setting the factors equal to zero:

1. \(x + 2 = 0 \) gives \(x = -2\)
2. \(x - 1 = 0 \) gives \(x = 1\)

Now we can test the intervals determined by these critical points: \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \).

- For \(x < -2\) (for example, \(x = -3\)):
\[
(-3 + 2)(-3 - 1) = (-1)(-4) = 4 > 0
\]

- For \(-2 < x < 1\) (for example, \(x = 0\)):
\[
(0 + 2)(0 - 1) = (2)(-1) = -2 < 0
\]

- For \(x > 1\) (for example, \(x = 2\)):
\[
(2 + 2)(2 - 1) = (4)(1) = 4 > 0
\]

From this testing, \( (x + 2)(x - 1) \leq 0 \) holds true in the interval \(-2 \leq x \leq 1\).

Thus, the solution to the inequality \(3x^2 + 3x - 6 \leq 0\) is:

\[
\boxed{-2 \leq x \leq 1}
\]