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Solve for xx\newline3(x+2)(x+5)>03(x+2)(x+5) > 0\newlineWrite the solution using interval notation. Use the union symbol ∪\cup to express the solution as a union of disjoint intervals. Finite endpoints of all intervals should be integers. If there are no solutions, use the symbol ∅\varnothing for the empty set. Use the set notation {a}\{a\} to represent an isolated solution aa. \newline□\square

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Q. Solve for xx\newline3(x+2)(x+5)>03(x+2)(x+5) > 0\newlineWrite the solution using interval notation. Use the union symbol ∪\cup to express the solution as a union of disjoint intervals. Finite endpoints of all intervals should be integers. If there are no solutions, use the symbol ∅\varnothing for the empty set. Use the set notation {a}\{a\} to represent an isolated solution aa. \newline□\square
  1. Set Inequality to Zero: Set the inequality 3(x+2)(x+5)>03(x+2)(x+5) > 0 to zero to find the critical points. This gives us the equations x+2=0x+2 = 0 and x+5=0x+5 = 0. Solving for xx gives us x=−2x = -2 and x=−5x = -5.
  2. Divide Number Line: Use the critical points to divide the number line into intervals: (−∞,−5),(−5,−2), and (−2,∞) (-\infty, -5), (-5, -2), \text{ and } (-2, \infty) .
  3. Test Intervals: Test a number from each interval in the inequality to determine where the inequality is true.\newlineFor (−∞,−5)(-\infty, -5), choose −6-6. Substituting −6-6 into the inequality gives 3(−6+2)(−6+5)=3(−4)(−1)=123(-6+2)(-6+5) = 3(-4)(-1) = 12, which is greater than 00.\newlineFor (−5,−2)(-5, -2), choose −3-3. Substituting −3-3 into the inequality gives 3(−3+2)(−3+5)=3(−1)(2)=−63(-3+2)(-3+5) = 3(-1)(2) = -6, which is not greater than 00.\newlineFor −6-600, choose 00. Substituting 00 into the inequality gives −6-633, which is greater than 00.
  4. Write Solution: Write the solution set in interval notation. The inequality is true for the intervals (−∞,−5)(-\infty, -5) and (−2,∞)(-2, \infty).\newlineThe solution in interval notation is (−∞,−5)∪(−2,∞)(-\infty, -5) \cup (-2, \infty).

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