$\underline{\text { Folklore }}$

Are there such values of $a$ and $b$ for which the equation $x^{4}-4 x^{3}+6 x^{2}+a x+b=0$ has four distinct real roots?
Let's write the equation in the form $(x-1)^{4}+(a+4) x+b-1=0$. After the substitution $t=x-1$, it will take the form: $t^{4}=-(a+4) t -(a+b+3)$.

The function $t^{4}$ is convex, so its graph cannot intersect the line $y=-(a+4) t-(a+b+3)$ at more than two points (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_388b3f8bbd0a17840215g-08.jpg?height=355&width=463&top_left_y=737&top_left_x=795)

## Answer

They do not exist.