# How do you find all the zeros of f(x)=2x^3+5x^2-2x-5?

Aug 5, 2016

$f \left(x\right)$ has zeros: $1$, $- 1$, $- \frac{5}{2}$

#### Explanation:

This cubic factors by grouping:

$f \left(x\right) = 2 {x}^{3} + 5 {x}^{2} - 2 x - 5$

$= \left(2 {x}^{3} + 5 {x}^{2}\right) - \left(2 x + 5\right)$

$= {x}^{2} \left(2 x + 5\right) - 1 \left(2 x + 5\right)$

$= \left({x}^{2} - 1\right) \left(2 x + 5\right)$

$= \left({x}^{2} - {1}^{2}\right) \left(2 x + 5\right)$

$= \left(x - 1\right) \left(x + 1\right) \left(2 x + 5\right)$

Hence zeros: $1$, $- 1$, $- \frac{5}{2}$