# How do you find all the zeros of f(x)=2x^3+5x^2+4x+1?

Aug 3, 2016

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

#### Explanation:

$f \left(x\right) = 2 {x}^{3} + 5 {x}^{2} + 4 x + 1$

Note that:

$f \left(- 1\right) = - 2 + 5 - 4 + 1 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

$2 {x}^{3} + 5 {x}^{2} + 4 x + 1$

$= \left(x + 1\right) \left(2 {x}^{2} + 3 x + 1\right)$

Substituting $x = - 1$ in the remaining quadratic, we find:

$2 {x}^{2} + 3 x + 1 = 2 - 3 + 1 = 0$

So $x = - 1$ is a zero again and $\left(x + 1\right)$ a factor:

$2 {x}^{2} + 3 x + 1 = \left(x + 1\right) \left(2 x + 1\right)$

The final zero is $x = - \frac{1}{2}$