A cubic function is given by \(f(x)=\left(2 x^2+3 x-5\right)(x+2)\)
- Find the zeros of \(f(x)\). (1 mark)
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- Hence, or otherwise, solve \(f(x) \geqslant 0\), giving your answer in set notation. (2 marks)
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Show Worked Solution
a. \(f(x)=\left(2 x^2+3 x-5\right)(x+2)=(2 x+5)(x-1)(x+2)\)
\(\text{Zeros at} \ \ x=-\dfrac{5}{2}, x=1 \ \ \text{and}\ \ x=-2\)
b. \(\text{Find} \ x \ \text{such that} \ \ f(x) \geqslant 0.\)
\(\text{At} \ \ x=0: \ (5)(-1)(2)<0\)
\(\therefore \text{Graph}\ (f(x)) \geqslant 0 \ \ \text{for} \ \ x \in\left[-\frac{5}{2},-2\right] \cup\ x \in[1, \infty)\)