Solve (3x+3)(2-x) > 0: Finding Positive Values of x

Look at the following function:

$y=\left(3x+3\right)\left(2-x\right)$

Determine for which values of $x$ the following is true:

$f(x) > 0$

To solve this problem, we'll follow these steps:

• Step 1: Find the roots of the function by setting each factor equal to zero.
• Step 2: Analyze the intervals determined by these roots.
• Step 3: Determine where the product of the factors is positive.

Now, let's work through each step:

Step 1: Find the roots of the function:
The function $y = (3x + 3)(2 - x)$ is zero when either $3x + 3 = 0$ or $2 - x = 0$.

Solving these equations:
$3x + 3 = 0 \Rightarrow x = -1$
$2 - x = 0 \Rightarrow x = 2$

Step 2: Analyze the intervals determined by the roots. The roots divide the number line into three intervals: $(-\infty, -1)$, $(-1, 2)$, and $(2, \infty)$.

Step 3: Determine the sign of $f(x)$ in each interval:

• For $x \in (-\infty, -1)$:
  Choose $x = -2$: $(3(-2) + 3)(2 - (-2)) = (-3)(4) = -12$. The product is negative.
• For $x \in (-1, 2)$:
  Choose $x = 0$: $(3(0) + 3)(2 - 0) = (3)(2) = 6$. The product is positive.
• For $x \in (2, \infty)$:
  Choose $x = 3$: $(3(3) + 3)(2 - 3) = (12)(-1) = -12$. The product is negative.

Therefore, the solution occurs when the product is positive, i.e., for values $x \in (-1, 2)$.

Thus, the intervals for which $f(x) > 0$ is $-2 < x < 1$.