# Write a simplified quartic equation with integer coefficients and positive leading coefficients as small as possible, whose single roots are -1/3 and 0 and has a double root as 0.4?

$75 {x}^{4} - 35 {x}^{3} - 8 {x}^{2} + 4 x = 0$

#### Explanation:

We have roots of:

$x = - \frac{1}{3} , 0 , \frac{2}{5} , \frac{2}{5}$

We can then say:

$x + \frac{1}{3} = 0 , x = 0 , x - \frac{2}{5} = 0 , x - \frac{2}{5} = 0$

And then:

$\left(x + \frac{1}{3}\right) \left(x\right) \left(x - \frac{2}{5}\right) \left(x - \frac{2}{5}\right) = 0$

And now starts the multiplying:

$\left({x}^{2} + \frac{1}{3} x\right) \left(x - \frac{2}{5}\right) \left(x - \frac{2}{5}\right) = 0$

$\left({x}^{2} + \frac{1}{3} x\right) \left({x}^{2} - \frac{4}{5} x + \frac{4}{25}\right) = 0$

${x}^{4} + \frac{1}{3} {x}^{3} - \frac{4}{5} {x}^{3} - \frac{4}{15} {x}^{2} + \frac{4}{25} {x}^{2} + \frac{4}{75} x = 0$

$75 {x}^{4} + 25 {x}^{3} - 60 {x}^{3} - 20 {x}^{2} + 12 {x}^{2} + 4 x = 0$

$75 {x}^{4} - 35 {x}^{3} - 8 {x}^{2} + 4 x = 0$