If $3a+2b+3c=0,$ show that the cubic polynomial $ax^3+bx+c$ has at least one root in the interval $(0,2)$

Question

In an old math forum (which was written about 15 years ago), I saw this question posed by a student and it had no answer. It intrigued me, so here it goes:

If $3a+2b+3c=0 \hspace{.1cm}(1),$ prove that the polynomial $p(x)=ax^3+bx+c$ has at least one root in the interval $(0,2).$

So, my first process of thought was: Let $$f(x)=\displaystyle \int p(x)dx=\frac{ax^4}{4}+\frac{bx^2}{2}+cx+d$$ so that $f'(x)=p(x)$. If I could prove now that, perhaps, $\exists \hspace{.1cm} x_1,x_2\in A\subseteq[0,2]$ such that $f(x_1)=f(x_2)$, by Rolle's Theorem, I would have that there is $r \in (x_1,x_2)\subseteq[0,2]$ such that $p(r)=0$ holds. However, I do not seem to be getting anywhere. I would very much appreciate your suggestions!

Answer 1

Notice that $a = -\frac{3c + 2b}{3}$ and hence $$a x^3 + bx + c = -\frac{3c + 2b}{3} x^3 + bx + c = c(1 - x^3) + b\left(x - \frac{2x^3}{3}\right)$$

If $b = 0$, I can take $x = 1$, and if $c = 0$, I can take $x = \sqrt{3}/\sqrt{2}$. So from now I can assume that $c\neq 0$ and $b\neq 0$. It is sufficient to show that $$\frac{1 - x^3}{x - \frac{2x^3}{3}} = -\frac{b}{c}$$ for some $x\in (0, 2)$. It follows from a fact that the image of the function $g(x) = \frac{1 - x^3}{x - \frac{2x^3}{3}}$ on $(0, 2)$ is $\mathbb{R}$. Indeed, $g(1) = 0$. When $x$ goes from $1$ to $\sqrt{3}/\sqrt{2}$, this function goes from $0$ to $-\infty$. In turn, when we go from $1$ to $0$, the function goes from $0$ to $+\infty$.

Answer 2

Wlog $a>0$

• If $c>0$ or $c<-a$ then $p(0)\cdot p(1) = -c(c+a)/2<0$ so we have a root in $(0,1)$
• If $c<0$ and $c>-a$ then $p(1)\cdot p(2) = (2c-5a)(c+a)/2<0$ so we have a root in $(1,2)$
• If $c=0$ then $p(x) = ax(2x^2-3)/2$ so $x = \sqrt{3/2}\in (0,2)$.
• If $c=-a$ then $b=0$ and $p(x) = a(x^3-1)$ so $x = 1\in (0,2)$.

Answer 3

Let $\lambda = a + b + c$ then $3 \lambda = b$ and $2 \lambda = - a - c$ so : $$p(x) = a x^3 + 3 \lambda x - (a + 2 \lambda)$$

1. If $a = 0$ then $p(x) = 3 \lambda x - 2 \lambda$.
   1. If $\lambda = 0$ then $p(1) = \lambda = 0$ and $1 \in (0, 2)$.
   2. If $\lambda \leq 0$ then $p \left(\dfrac{2}{3}\right) = 0$ and $\dfrac{2}{3} \in (0, 2)$.
2. If $a \neq 0$ then $p$ and $q = \dfrac{p}{a}$ have same roots. Let $\alpha = \dfrac{\lambda}{a}$ then : $$q(x) = x^3 + 3 \alpha x - (1 + 2 \alpha)$$
   1. If $\alpha < -\dfrac{1}{2}$ then $q(1) = \alpha < 0$ and $q(0) = -(1 + 2 \alpha) > 0$ so $\exists x \in (0,1) \subset (0, 2), q(x) = 0$.
   If $\alpha > -\dfrac{1}{2}$ then $q(0) = -(1 + 2 \alpha) < 0$ and $q(2) = 7 + 4 \alpha > 7 - 2 > 0$ so $\exists x \in (0,2), q(x) = 0$.
   2. If $\alpha = -\dfrac{1}{2}$ then : $$q(x) = x^3 - \dfrac{3}{2} x = x \left(x^2 - \dfrac{3}{2}\right)$$ and we deduce that $\sqrt{\dfrac{3}{2}}$ is a root of $q$ in $(0, 2)$.