How do you find all the zeros of f(x)=2x^3+x^2-8x-4?

Mar 15, 2016

Zeros of $f \left(x\right)$ are $\left\{2 , - 2 , - \frac{1}{2}\right\}$

Explanation:

To find zeros of $f \left(x\right) = 2 {x}^{3} + {x}^{2} - 8 x - 4$, first identify factors $- 4$, for which $f \left(x\right) = 0$. If its zero for $x = a$, $\left(x - a\right)$ will be a factor of $f \left(x\right)$.

Factors of $- 4$ are $\left\{1 , - 1 , 2 , - 2 , 4 , - 4\right\}$ and it is apparent that for $x = 2$, $f \left(2\right) = 2 {\left(2\right)}^{3} + {2}^{2} - 8 \cdot 2 - 4 = 16 + 4 - 16 - 4 = 0$, hence $\left(x - 2\right)$ is a factor of $2 {x}^{3} + {x}^{2} - 8 x - 4$. Dividing latter by former

$2 {x}^{3} + {x}^{2} - 8 x - 4 = \left(x - 2\right) \left(2 {x}^{2} + 5 x + 2\right)$

Now $\left(2 {x}^{2} + 5 x + 2\right)$ can be easily factorized by splitting $5 x$ as $4 x + x$ - (In a quadratic polynomial $a {x}^{2} + b x + c$, one has to identify factors of $a c$, whose sum is $b$. Hence, here factors of $2 \times 2 = 4$ are $4$and $1$, whose sum is $5$).

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

Hence, factors of $f \left(x\right) = 2 {x}^{3} + {x}^{2} - 8 x - 4$ are $\left(x - 2\right) \left(x + 2\right) \left(2 x + 1\right)$.

As zeros of $f \left(x\right)$ are given by $x - 2 = 0$, $x + 2 = 0$ and $2 x + 1 = 0$

Hence, zeros of $f \left(x\right)$ are $\left\{2 , - 2 , - \frac{1}{2}\right\}$