# $n$th degree polynomials $P(x) = Q(x)P''(x)$ with $Q$ quadratic, if $P$ has $\ge 2$ distinct roots then then $n$ distinct roots.

Let $P(x)$ be a polynomial of degree $n$ such that $P(x) = Q(x)P''(x)$ for some quadratic polynomial $Q$. Show that if $P$ has at least two distinct roots then it must have $n$ distinct roots.

• What do you think, Obama? – miradulo Sep 27 '15 at 21:19

Suppose otherwise; then we may write$$P(x) = (x-r)^\ell H(x)$$for some $r\in\mathbb{C}$, integer $\ell\ge2$, and nonconstant $H$ with $H(r)\ne0$. The condition now translates to$$(x-r)^\ell H(x) = Q(x)((x-r)^\ell H''(x) + 2\ell(x-r)^{\ell-1}H'(x) + \ell(\ell-1)(x-r)^{\ell-2}H(x)),$$ or $$(x-r)^2 H(x) = Q(x)((x-r)^2 H''(x) + 2\ell(x-r)H'(x)+\ell(\ell-1)H(x)).$$But $$\ell(\ell-1)H(r)\ne0,$$so$$(x-r)^2\mid Q(x)\implies Q(x) = c(x-r)^2$$for some $c\ne0$; hence$$H(x) = c(\ell(\ell-1)H(x)+2\ell(x-r)H'(x)+(x-r)^2 H''(x)).$$ Plugging in $x=r$ yields$$1 = c\ell(\ell-1),$$so$$2\ell H'(x)+(x-r)H''(x) = 0.$$It follows that$$\log H'(x) = -2\ell\log(x-r) + d$$for some constant $d$ (and all sufficiently large $x$), so$$H'(x) = e^d(x-r)^{-2\ell}\implies H'(x)(x-r)^{2\ell} = e^d$$for infinitely many $x$, which is absurd.