Determining multiplicity of a root of a polynomial using derivatives (or otherwise?)

Solution 1:

The theorem is

If $P(x)$ has a zero of order $k > 0$ at $x=x_0$, then $P'(x)$ has a zero of order $k-1$ at $x=x_0$.

As a consequence,

$P(x)$ has a zero of order $k \geq 0$ at $x=x_0$ if and only if $$P(x_0) = P'(x_0) = \cdots = P^{(k-1)}(x_0) = 0$$ and $P^{(k)}(x_0) \neq 0$.

Here, we have $P_n(x) = nx^{n+2} - (n+2)x^{n+1} + (n+2)x - n$ and $$P_n(1) = n - (n+2) + (n+2) - n = 0 \\ P'_n(1) = n(n+2) - (n+2)(n+1) + (n+2) = 0 \\ P''_n(1) = n(n+2)(n+1) - (n+2)(n+1)n = 0 \\ P'''_n(1) = n(n+2)(n+1)n - (n+2)(n+1)n(n-1) = (n+2)(n+1)n \neq 0$$ so $P_n(x)$ has a zero of order $3$ at $x=1$ for all $n > 0$.

You mentioned a theorem using indefinite integrals and integration constants, but this is effectively doomed to failure since integration constants are not intrinsic to the function but rather are artifacts of the method used to integrate. The closest we may get is

If $P(x)$ has a zero of order $k \geq 0$ at $x=x_0$, then $\int_{x_0}^x P(t)\,dt$ has a zero of order $k+1$ at $x=x_0$.

but I don't see what properties this theorem would yield that the above theorem doesn't already grant.

Solution 2:

No, the answer is not $n+2$ in general. Take, say, $n=6$. Then the polynomial factorizes as $$ 2(3x^4 + 2x^3 + 4x^2 + 2x + 3)(x + 1)(x - 1)^3, $$ so the multiplicity of the root $1$ is $3$ in this case. The same is true for all $n$ I have tested.

Solution 3:

There is one thing which could be interesting to do. Let $x=y+1$ to make $$2+(n+2)y+(y+1)^{n+1} (n y-2)=0$$ Now, use the binomial theorem or Taylor expansion around $y=0$ to get $$\frac{n(n+1)(n+2)}{6} y^3+\frac{(n-1)n(n+1)(n+2)}{12} y^4+O\left(y^5\right)$$ So, for an integer value of $n$, the multiplicity is $3$ (as @Dietrich Burde already reported).

Solution 4:

The following proves that $\,(x-1)^3 \,\mid\, P_{n+1}(x)-xP_n(x)\,$, then given that $\,P_1(x)=(x-1)^3\,$ it follows by induction that $\,(x-1)^3 \,\mid\, P_n(x)\,$ for all $\,n\,$.

$$ \require{cancel} \begin{align} P_{n+1}(x)-xP_n(x) &= x^{n+3}-x^{n+2}-(n+2)x^2+(2n+3)x-n-1 \\ &= x^{n+2}(x-1)-((n+2)x-n-1)(x-1) \\ &= (x-1)(x^{n+2}-(n+2)x+n+1) \\ &= (x-1)(x^{n+2}-1-(n+2)(x-1)) \\ &= (x-1)^2(x^{n+1}+x^{n}+\dots+x+1-(n+2)) \\ &= (x-1)^2 \big(\underbrace{(x^{n+1}-1)}_{=\,(x-1)(\cdots)}+\underbrace{(x^n-1)}+\dots+\underbrace{(x-1)}+(\cancel{1}-\cancel{1})\big) \\ &= (x-1)^3(\;\cdots\;) \end{align} $$

[ EDIT ] $\;$ For an alternative way, $\,P_n(1)=0\,$ so it is enough to show that $\,(x-1)^2 \,\mid P_n^{'}(x)\,$. But $\,P_n^{'}(x)=(n+2)\big(nx^{n+1}-(n+1)x^n + 1\big)\,$, answered at Show that there exist a polynomial $q_n(x) \in \mathbb Q[x] $ satisfying $p_n(x)=(x-1)^2q_n(x)$, which also explains that déjà vu feeling ;-)