Solve:
$\begin{array}{l}
(x+1)\left(x^2-5 x+6\right) \leq 0 \\{[?] \leq x \leq \quad \text { or } x \leq\end{array}$
Answer (2)
Factor the quadratic: x 2 − 5 x + 6 = ( x − 2 ) ( x − 3 ) .
Rewrite the inequality: ( x + 1 ) ( x − 2 ) ( x − 3 ) ≤ 0 .
Find critical points: x = − 1 , 2 , 3 .
Determine intervals: x ≤ − 1 or 2 ≤ x ≤ 3 . The final answer is 2 ≤ x ≤ 3 or x ≤ − 1 .
Explanation
Factor the quadratic expression First, we need to factor the quadratic expression x 2 − 5 x + 6 . We are looking for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. Therefore, we can factor the quadratic as ( x − 2 ) ( x − 3 ) .
Rewrite the inequality Now we can rewrite the inequality as ( x + 1 ) ( x − 2 ) ( x − 3 ) ≤ 0 .
Find the critical points Next, we find the critical points by setting each factor to zero and solving for x . The critical points are x = − 1 , x = 2 , and x = 3 .
Create a sign chart Now we create a sign chart to determine the intervals where the inequality is satisfied. We will test the intervals ( − ∞ , − 1 ) , ( − 1 , 2 ) , ( 2 , 3 ) , and ( 3 , ∞ ) .
For x < − 1 , let's test x = − 2 . Then ( x + 1 ) ( x − 2 ) ( x − 3 ) = ( − 2 + 1 ) ( − 2 − 2 ) ( − 2 − 3 ) = ( − 1 ) ( − 4 ) ( − 5 ) = − 20 , which is less than or equal to 0. So, the interval ( − ∞ , − 1 ] is part of the solution.
For − 1 < x < 2 , let's test x = 0 . Then ( x + 1 ) ( x − 2 ) ( x − 3 ) = ( 0 + 1 ) ( 0 − 2 ) ( 0 − 3 ) = ( 1 ) ( − 2 ) ( − 3 ) = 6 , which is greater than 0. So, the interval ( − 1 , 2 ) is not part of the solution.
For 2 < x < 3 , let's test x = 2.5 . Then ( x + 1 ) ( x − 2 ) ( x − 3 ) = ( 2.5 + 1 ) ( 2.5 − 2 ) ( 2.5 − 3 ) = ( 3.5 ) ( 0.5 ) ( − 0.5 ) = − 0.875 , which is less than or equal to 0. So, the interval [ 2 , 3 ] is part of the solution.
For 3"> x > 3 , let's test x = 4 . Then ( x + 1 ) ( x − 2 ) ( x − 3 ) = ( 4 + 1 ) ( 4 − 2 ) ( 4 − 3 ) = ( 5 ) ( 2 ) ( 1 ) = 10 , which is greater than 0. So, the interval ( 3 , ∞ ) is not part of the solution.
Identify the intervals The intervals where the expression ( x + 1 ) ( x − 2 ) ( x − 3 ) is less than or equal to zero are ( − ∞ , − 1 ] and [ 2 , 3 ] . Therefore, the solution set is x ≤ − 1 or 2 ≤ x ≤ 3 .
Final Answer Thus, the solution to the inequality ( x + 1 ) ( x 2 − 5 x + 6 ) ≤ 0 is 2 ≤ x ≤ 3 or x ≤ − 1 .
Examples
Understanding inequalities like this is crucial in many real-world scenarios. For instance, imagine you're designing a bridge. The materials used must withstand certain stress levels. This stress can be modeled by an inequality, ensuring the bridge's stability. Similarly, in economics, inequalities help determine price ranges for products to ensure profitability while remaining competitive. These mathematical tools provide a framework for making informed decisions in engineering, economics, and beyond.
The solution to the inequality ( x + 1 ) ( x 2 − 5 x + 6 ) ≤ 0 is found by factoring and determining intervals. The final answer is x ≤ − 1 or 2 ≤ x ≤ 3 .
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