What should $a, b \in\Bbb{R}$ be that polinom $x^5+ax^3+b\in\Bbb{R}[x]$ would have at least one non-zero repeated root. What should $a, b\in\Bbb{R}$ be that polinom $x^5+ax^3+b\in\Bbb{R}[x]$ would have at least one non-zero repeated root.
Well, what I found out by myself are there: $$1)\quad  b=0,\quad then\quad
x^5+ax^3=x^3(x^2+a),\quad a\in\Bbb{R}$$
$$2) \quad a=0,\quad then\quad x^5+b=0, \quad x=-b^{\frac{1}{5}}, \quad b=n^5,\quad n\in\Bbb{R}$$
But these are only several variants...
 A: Let $f(x)=x^5+ax^3+b$. If $\phi$ is a repeated root, then $f(\phi)=f'(\phi)=0$.
$$f'(\phi)=5\phi^4+3a\phi^2$$
$$\phi f'(\phi)-5f(\phi)=0$$
$$5\phi^5+3a\phi^3-5\phi^5-5a\phi^3-5b=0$$
$$-2a\phi^3-5b=0$$
So $f(\phi)=(\phi^2+a)(\frac{-5b}{2a})+b=0$.
$$\phi^2=\frac{-3a}{5} \quad\text{or}\quad b=0$$
$b=0$ is an answer.
If $b\ne0$, then
$$\left(\frac{-3a}{5}\right)^3=\phi^6=\left(\frac{-5b}{2a}\right)^2$$
That is $108a^5+3125b^2=0$.
A: The polynomial $x^5+ax^3+b$ has (at least) a repeated root iff its discriminant is zero, i.e. iff $b=0$ or $108 a^5 + 3125 b^2=0$.
A: Hints:
If $\;f(x)=x^5+ax^3+b\;$ , then a root is multiple iff it is also a root of the polynomial's derivative:
$$f'(x)=5x^4+3ax^2=x^2(5x^2+3a)$$
so if $\;b=0\;$ then $\;x=0\;$ is a multiple root, otherwise , if $\;w\neq0\;$ is a multiple root, it must be that both
$$\begin{cases}w^5+aw^3+b=0\implies&a=-\cfrac{w^5+b}{w^3}\\{}\\
5w^2+3a=0\implies&a=-\cfrac{5w^2}3\end{cases}\;\;\;\implies 3w^5+3b=5w^5\implies$$$${}$$
$$\implies2w^5=3b\implies w=\sqrt[5]\frac{3b}2$$
Try now to find out the conditions on the coefficients.
A: The discriminant of your polynomial is $b^2 (108 a^5 + 3125 b^2)$.  There is a repeated root iff this is $0$.  That occurs of course for $b=0$, but then the root in question is $0$.  Otherwise, you want $a = (-3125 b^2/108)^{1/5}$.
