Conditions to have a triple root

Question

Find the condition such that the polynomial $x^5+10ax^3+5bx+c $ has a triple root which is different from zero where $x \in\mathbb{R}$. Do not use the discriminant.

Answer 1

We have$$f^{\prime}(x)=5x^4+30ax^2+5b,\quad f^{\prime\prime}(x)=20x^3+60ax.$$ If $x$ is a root at least triple and non null of $f(x),$ $$\left \{ \begin{matrix} x^5+10ax^3+5bx+c=0 \\5x^4+30ax^2+5b=0 \\20x^3+60ax=0\end{matrix}\right.\underbrace{\Leftrightarrow}_{x\neq 0} \left \{ \begin{matrix} x^5+10ax^3+5bx+c=0 \\x^4+6ax^2+b=0 \\x^2+3a=0.\end{matrix}\right.$$ Substituting $x^2=-3a$ in the second equality $$9a^2-18a^2+b=0, \text{ or }b=9a^2.$$ The first equality is equivalent to $x\left(x^4+10ax^2+5b\right)+c=0.$ Substituting in this $x=\sqrt{-3a},$ $$\sqrt{-3a}\left(9a^2-30a^2-b\right)=-c\text{ or }24a^2\sqrt{-3a}=-c.$$ Taking squares $-1728a^5=c^2.$ So, the asked conditions are$$\boxed{\left \{ \begin{matrix} b=9a^2\\1728a^5+c^2=0.\end{matrix}\right.}$$

Answer 2

If it has a triple root, it factors as $$f(x)=(x-r)^3q(x)$$ where $q(x)$ is quadratic. If you differentiate it twice, you have $$f'(x)=3(x-r)^2q(x)+(x-r)^3q'(x)$$ $$f''(x)=6(x-r)q(x)+6(x-r)^2q'(x)+(x-r)^3q''(x)$$

And you should note that $r$ is not only a root of $f$, but also of $f'$ and $f''$. So the gcd of the three polynomials is divisible by $(x-r)$. That is, they all have a root $r$ in common. So $$f(x)=x^5+10ax^3+5bx+c$$ $$f'(x)=5x^4+30ax^2+5b$$ $$f''(x)=20x^3+60ax=20x(x^2+3a)$$ have the root $r$ in common. It is easy to see what are the roots of $f''$, and one of them is $0$ which you are supposed to not consider. After writing down the other two roots, what would it take for one of these to also end up being a root of $f'$?This gives you a condition on how $b$ relates to $a$. Then with that information, what would it take for the root to be a root of $f$? This in turn gives you a condition for $c$ in terms of $a$.

Ultimately, I think if $a$ is any negative number, you end uf with formulas for $b$ and $c$ in terms of $a$ that give it the triple root.

Answer 3

\begin{align*} f(x) &= x^5+10ax^3+5bx+c \\ f'(x) &= 5x^4+30ax^2+5b \\ f''(x) &= 20x^3+60ax \end{align*}

For triple root,

$$f(x)=f'(x)=f''(x)=0$$

where $x\ne 0$

\begin{align*} 20x^3+60ax &= 0 \\ x(x^2+3a) &= 0 \\ x^2 &= -3a \\ x &= \color{red}{\boldsymbol{\pm}} \sqrt{-3a} \\ 5(-3a)^2+30a(-3a)+5b &= 0 \\ b &= 9a^2 \\ x^5+10ax^3+5(9a^2)x+c &= 0 \\ c &= -x[(-3a)^2+10a(-3a)+45a^2] \\ &= -24a^2x \\ &= \color{blue}{\boldsymbol{\mp}} 24a^2 \sqrt{-3a} \\ f(x) &= x^5+10ax^3+45a^2x \color{blue}{\boldsymbol{\mp}} 24a^2\sqrt{-3a} \\ &= (x \color{blue}{\boldsymbol{\mp}} \sqrt{-3a})^{3} (x^2 \color{red}{\boldsymbol{\pm}} 3\sqrt{-3a}\, x-8a) \end{align*}

Answer 4

The polynomial should be divisible by $(x-u)^3$ for some $u$. Long division gives a remainder $$ (10u^3+30au)x^2+ (5b-30au^2-15u^4)x+ (c+10au^3+6u^5) $$ which should be zero.