Question

How to solve (x+9)(x+3)(x-2)<0solution in interval notation

Answer 1

`(-\infty\text{ -9) }\cup\text{ (-3, 2)}`

Step-by-step explanation:
`\mleft(x+9\mright)\mleft(x+3\mright)\mleft(x-2\mright)<0`

To solve the inequality, we need to find the values of x

`\begin{gathered} \text{When we have quadratic equatio written as (x - a)(x - b) = 0} \\ We\text{ solve by seperating them individually: }x\text{ - a = 0 or x - b = 0 } \\ \text{We will apply same here:} \\ (x\text{ +9) < 0 or (x + 3) < 0 or (x - 2) < 0} \\ (x\text{ +9) < 0} \\ \text{subtract 9 to both sides:} \\ x\text{ + 9 - 9 < 0 - 9} \\ x\text{ < -9} \end{gathered}`

`\begin{gathered} (x\text{ + 3) < 0} \\ subtract\text{ 3 from both sides:} \\ x\text{ + 3 - 3 < 0 - 3} \\ x\text{ < -3} \\ \\ (x\text{ - 2) < 0} \\ \text{add 2 to both sides:} \\ x\text{ - 2 + 2 < 0 + 2} \\ x\text{ < 2} \end{gathered}`

`\begin{gathered} x\text{ <-9} \\ (-\infty,\text{ -9)} \\ \text{for x < -3 and x < 2} \\ -3<\text{ x< 2} \\ (-3,2) \\ \text{The solution becomes:} \\ (-\infty\text{ -9) }\cup\text{ (-3, 2)} \end{gathered}`