# How do you find all the zeros of f(x) = x^4 - 3.3x^3 + 2.3x^2 + 0.6x?

May 5, 2016

Zeros of $f \left(x\right)$ are $\left\{0 , 2 , \frac{3}{2} , - \frac{1}{5}\right\}$

#### Explanation:

$f \left(x\right) = {x}^{4} - 3.3 {x}^{3} + 2.3 {x}^{2} + 0.6 x$

= $\frac{x}{10} \left(10 {x}^{3} - 33 {x}^{2} + 23 x + 6\right)$

Hence one of the zeros is $0$. Another zero could be a factor of $6$

As putting $x = 2$ in $\left(10 {x}^{3} - 33 {x}^{2} + 23 x + 6\right)$, we get

$10 \cdot {2}^{3} - 33 \cdot {2}^{2} + 23 \cdot 2 + 6 = 80 - 132 + 46 + 6 = 0$

Hence, $2$ is another zero and dividing $10 {x}^{3} - 33 {x}^{2} + 23 x + 6$ by $\left(x - 2\right)$, we get

$10 {x}^{3} - 33 {x}^{2} + 23 x + 6 = \left(x - 2\right) \left(10 {x}^{2} - 13 x - 3\right)$

or $\left(x - 2\right) \left(10 {x}^{2} - 15 x + 2 x - 3\right)$

or $\left(x - 2\right) \left(5 x \left(2 x - 3\right) + 1 \cdot \left(2 x - 3\right)\right)$

= $\left(x - 2\right) \left(5 x + 1\right) \left(2 x - 3\right)$

Hence other zeros are given by $2 x - 3 = 0$ and $5 x + 1 = 0$ i.e. are $\frac{3}{2}$ and $- \frac{1}{5}$

Hence zeros of $f \left(x\right)$ are $\left\{0 , 2 , \frac{3}{2} , - \frac{1}{5}\right\}$