Solve for x. (x - 2)(x - 3) > 0 Write a compound inequality like 1 3. ______

Find Zeros: Find the zeros of the quadratic expression (x - 2)(x - 3). (x - 2) = 0 gives x = 2. (x - 3) = 0 gives x = 3. Critical points are 2 and 3.Determine Intervals: Determine the intervals to test using the critical points. The intervals are (-∞, 2), (2, 3), and (3, ∞).Test Interval (-∞, 2): Test the interval (-∞, 2) by picking a number less than 2, say x = 0. (0 - 2)(0 - 3) = (2)(3) = 6, which is > 0. So, (x - 2)(x - 3) > 0 for x in (-∞, 2).Test Interval (2, 3): Test the interval (2, 3) by picking a number between 2 and 3, say x = 2.5. (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25, which is 0 for x in (2, 3).Test Interval (3, ∞): Test the interval (3, ∞) by picking a number greater than 3, say x = 4. (4 - 2)(4 - 3) = (2)(1) = 2, which is > 0. So, (x - 2)(x - 3) > 0 for x in (3, ∞).Combine Intervals: Combine the intervals where (x - 2)(x - 3) > 0. The solution is x in (-∞, 2) or (3, ∞).

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Solve for $x$ . $\newline$ (x - 2)(x - 3) > 0 $\newline$ Write a compound inequality like 1 < x < 3 or like x < 1 or x > 3.

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Q. Solve for

x

\newline

(x - 2)(x - 3) > 0

\newline

Write a compound inequality like

1 < x < 3

or like

x < 1

x > 3

Find Zeros: Find the zeros of the quadratic expression $(x - 2)(x - 3)$ . $(x - 2) = 0$ gives $x = 2$ . $(x - 3) = 0$ gives $x = 3$ .Critical points are $2$ and $3$ .
Determine Intervals: Determine the intervals to test using the critical points. $\newline$ The intervals are $(-\infty, 2)$ , $(2, 3)$ , and $(3, \infty)$ .
Test Interval $(-\infty, 2):$ Test the interval $(-\infty, 2)$ by picking a number less than $2$ , say $x = 0$ . $(0 - 2)(0 - 3) = (2)(3) = 6$ , which is < 0. So, (x - 2)(x - 3) > 0 for $x$ in $(-\infty, 2)$ .
Test Interval $(2, 3)$ : Test the interval $(2, 3)$ by picking a number between $2$ and $3$ , say $x = 2.5$ . $(2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25$ , which is < 0. So, $(x - 2)(x - 3)$ is not > 0 for $x$ in $(2, 3)$ .
Test Interval $(3, \infty):</b> Test the interval \$(3, \infty)$ by picking a number greater than $3$ , say $x = 4$ . $(4 - 2)(4 - 3) = (2)(1) = 2$ , which is > 0. $\newline$ So, (x - 2)(x - 3) > 0 for $x$ in $(3, \infty)$ .
Combine Intervals: Combine the intervals where (x - 2)(x - 3) > 0. The solution is $x$ in $(-\infty, 2)$ or $(3, \infty)$ .

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Solve for xxx.\newline(x - 2)(x - 3) > 0 \newlineWrite a compound inequality like 1 < x < 3 or like x < 1 or x > 3.

Solve for $x$ . $\newline$ (x - 2)(x - 3) > 0 $\newline$ Write a compound inequality like 1 < x < 3 or like x < 1 or x > 3.