Polynomial with rational roots

Question

Let $f:\mathbb{R} \to \mathbb{R}$ be a polynomial function with rational coefficients and degree $3$. If the graphic of $f$ is tangent to the $x$ axis. Show that all zeros of $f$ are rational numbers.

If $\alpha$ is a zero of $f$, i.e., if $f(\alpha)=0$ and the graph of $f$ is tangent to the $x$ axis in this point, then $\alpha$ has even multiplicity, but I don't know how I can conclude this question. Someone can help-me?

Thanks a lot.

Answer 1

Hint Since $$f(x) = A x^3 + B x^2 + C x + D$$ is rational and scaling a polynomial doesn't change its roots, we can replace $f$ with $A^{-1} f$ and hence assume it is monic: $$f(x) = x^3 + b x^2 + c x + d .$$ Similarly, $f(x)$ has rational roots iff $f(x - \frac{b}{3})$ does, so we may as well prove the claim for $f(x - \frac{b}{3})$, but expanding gives that $f(x - \frac{b}{3})$ is rational and has zero quadratic coefficient, so we may as well assume $f(x)$ has the form $$f(x) = x^3 + p x + q$$ for rational $p, q$ (we call such a cubic depressed).

Now, since the root $\alpha$ has multiplicity $\geq 2$, $f$ must have the form $$f(x) = (x - \alpha)^2 (x - \beta) .$$ Expand and compare like terms in $x$.

Doing so gives $\beta = -2 \alpha$ (so that $\beta$ is rational iff $\alpha$ is) and thus $$p = -3 \alpha^2, q = 2 \alpha^3.$$ Thus, either $\alpha = 0$ or (if $p \neq 0$) $$\alpha = -\frac{3 q}{2 p}$$ and either way $\alpha$ is rational. (If we like, we can back-substitute to find explicit rational expressions for $\alpha, \beta$ in terms of $A, B, C, D$.)

Answer 2

If $\alpha$ is a root where the graph of $f$ is tangent to the $x$ axis, then $\alpha$ is a root of multiplicity $\ge 2$ (not necessarily even).

Case 1: $\alpha$ has multiplicity $3$. Then $\alpha$ is a root of the polynomial $f''(x)$ which has degree $1$ and rational coefficients, and therefore $\alpha$ is rational.

Case 2: $\alpha$ has multiplicity $2$. Then the gcd of $f(x)$ and $f'(x)$ is $x-\alpha$. But the gcd of two polynomials with rational coefficients has rational coefficients, so $\alpha$ must be rational.