For which values of \(x\) is \((2x^2-3x-5)(x+2)<0\) ?
- \(x \in(-1,-2) \ \cup\ x \in\left(\dfrac{5}{2}, \infty\right]\)
- \(x \in[-\infty,-2)\ \cup\ x \in\left(-1, \dfrac{5}{2}\right)\)
- \(x \in(-1,-2)\ \cup\ x \in\left(\dfrac{5}{2}, \infty\right)\)
- \(x \in(-\infty,-2)\ \cup\ x \in\left(-1, \dfrac{5}{2}\right)\)
Show Worked Solution
\((2x^2-3x-5)(x+2)=(2x-5)(x+1)(x+2) \)
\(\text {Zeros at} \ \ x=\dfrac{5}{2}, \ x=-1,\ \ \text {and} \ \ x=-2\)
\(\text{At} \ \ x=0: \ (-5)(1)(2)<0\)
\(\therefore \ \text{Graph }<0\ \text{ for }\ x \in(-\infty,-2)\ \cup\ x \in\left(-1, \frac{5}{2}\right) \)
\(\Rightarrow D\)