Multiplicity of root $1$ of $nX^{n+2}-(n+2)X^{n+1}+(n+2)X-n$?

Question

I'm solving this homework problem:

Show that $1$ is a root of the polynomial $nX^{n+2}-(n+2)X^{n+1}+(n+2)X-n$. Determine its multiplicity.

I believe the answer is $n+2,$ and currently working on proving this. My intuition stems from the fact that when $n=1, P(x)=(x-1)^3, so$ 1 is a polynomial of multiplicity $3.$

I attempted this: calling the polynomial $P(x),$ we see that $P(1)=0,$ hence $1$ is a root for sure. To determine its multiplicity, I'm thinking about:

Taking successive derivatives $P'(1), P''(1) \dots P^{(n+3)}(1) $ of the polynomial and showing that all but the last one $P^{(n+3)}(1)$ vanishes. But I'm a bit skeptical because the antiderivative of a polynomial doesn't always have multiplicity one more than the original polynomial, which forms the basis/idea of the method I'm thinking of. For example $Q(x):=2x$ has the root $0$ of multiplicity $1,$ but the antiderivative $x^2+1$ doesn't have any real root.

So my question is: in order to answer the question in the image, what is the exact theorem should we use? Is it something like this?

Proposed theorem: If $Q(x)$ has the real root $a$ of order $k,$ then its antiderivative with constant of integration $0$ has the real root $a$ of order $k+1.$

The above seems to be true, and if yes, can we use this to show that the successive derivatives $P'(1), P''(1) \dots P^{(n+2)}(1) $ of the polynomial vanish but $P^{(n+2)}(1) $ doesn't vanish, and this'll show that $1$ is a root of multiplicity $n+2.$ Is my idea correct?

ADDENDUM/EDIT: Can we arrive at the proof that $1$ is a root of multiplicity $n+2$ by using induction? At $n=1,$ the multiplicity is $1+2=3,$ so the induction can start, and now we just need to show the induction step. Will this work? P.S. As per Dietrich's answer, the multiplicity seems to be $3$ irrespective of $n.$ So can we just prove this using the derivatives above or by induction?

Answer 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 $n > 0$.

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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.

Answer 2

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).

Answer 3

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.

Answer 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 $\,n\,$.

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

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[ 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 ;-)

Answer 5

We may also look at what the coefficients tell us: the polynomial $$ \ p(x) \ \ = \ \ n·x^{n+2} \ - \ (n+2)·x^{n+1} \ + \ 0·x^n \ + \ \ldots \ + \ 0·x^2 \ + \ (n+2)·x \ - \ n \ \ $$ is anti-palindromic, which is to say that the coefficients "read left-to-right" are the negative of the sequence "read right-to-left". Consequently, $ \ x \ = \ 1 \ $ must be a zero of such a polynomial, as the terms in $ \ p(1) \ $ then "cancel in pairs". We also see from the Rule of Signs that $ \ p(x) \ $ has three "sign-changes", indicating that it has either three or one positive real zeroes, regardless of the parity of $ \ n \ \ . $ Since for $ \ n \ $ even, $$ \ p(-x) \ \ = \ \ n·x^{n+2} \ + \ (n+2)·x^{n+1} \ + \ 0·x^n \ + \ \ldots \ + \ 0·x^2 \ - \ (n+2)·x \ - \ n \ \ $$ has one "sign-change", there is also one negative real root; for $ \ n \ $ odd, $$ \ p(-x) \ \ = \ \ -n·x^{n+2} \ - \ (n+2)·x^{n+1} \ + \ 0·x^n \ + \ \ldots \ + \ 0·x^2 \ - \ (n+2)·x \ - \ n \ \ $$ has no "sign-changes", so in this case, $ \ p(x) \ $ has no negative real roots.

Upon employing polynomial/synthetic division for $ \ x \ = \ 1 \ \ , $ we obtain the "reduced" polynomial $$ p_1(x) \ \ = \ \ n·x^{n+1} \ \underbrace{- \ 2·x^n \ - \ 2·x^{n-1} \ + \ \ldots \ - \ 2·x^2 \ - \ 2·x}_{n \ \text{terms}} \ + \ n \ \ , $$ which is palindromic, thus having the property that if $ \ r \ $ is a zero, then $ \ \frac{1}{r} \ $ is also. We observe that $ \ x \ = \ 1 \ $ is a zero of this polynomial as well, so it also has a zero $ \ x \ = \ \frac11 \ = \ 1 \ \ , \ $ which establishes that the three positive real zeroes of $ \ p(x) \ $ are $ \ 1 \ $ with multiplicity $ \ 3 \ \ . $

With $ \ n \ $ odd then, there are these three real zeroes with the remaining $ \ (n - 1) \ $ [even] zeroes forming complex-conjugate pairs. For $ \ n \ $ even, the anti-palindromic character of $ \ p(x) \ $ implies that $ \ x \ = \ -1 \ $ is also a zero, giving us the expected three positive and one negative real zeroes and $ \ (n - 2) \ $ [even] complex-conjugate zeroes.

It might be mentioned, by Viete's relations, that as the product of all $ \ (n + 2) \ $ zeroes of $ \ p(x) \ $ is $ \ +1 \ $ for $ \ n \ $ odd and $ \ -1 \ $ for $ \ n \ $ even, the product of the complex zeroes alone is always $ \ +1 \ \ . $ This means that the complex conjugate zeroes are in pairs $ \ r \ , \ \overline{r} \ = \ \frac{1}{r} \ \ , $ so all of the complex zeroes have unit modulus (so in fact all of the zeroes do).

We can also demonstrate the triple multiplicity by polynomial division. Dividing $ \ p_1(x) \ $ by $ \ (x - 1) \ $ produces $$ p_2(x) \ \ = \ \ n·x^n \ + \ \underbrace{(n - 2)·x^{n-1} \ + \ (n - 4)·x^{n-2} \ + \ \ldots \ + \ (n - [2n - 2] )·x \ + \ (n - 2n)}_{n \ \text{terms}} $$ $$ = \ \ n·x^n \ + \ (n - 2)·x^{n-1} \ + \ (n - 4)·x^{n-2} \ + \ \ldots \ - \ (n - 4)·x^2 \ - \ (n -2)·x \ - \ n \ \ , $$ which is anti-palindromic, and so has $ \ x \ = \ 1 \ $ as a zero. One further division by $ \ (x - 1) \ $ yields $$ p_3(x) \ \ = \ \ n·x^{n-1} \ + \ \underbrace{2·(n - 1)·x^{n-2} \ + \ 3·(n - 2)·x^{n-3} \ + \ \ldots \ + \ (n - 1)·(n - [n - 2] )·x \ + \ n·(n - [n - 1])}_{(n - 1) \ \text{terms}} $$ $$ = \ \ n·x^{n - 1} \ + \ 2·(n - 1)·x^{n-2} \ + \ 3·(n - 2)·x^{n-3} \ + \ \ldots $$ $$ + \ 3·(n - 2)·x^2 \ + \ 2·(n - 1)·x \ + \ n \ \ . $$ Since the exponents in the terms are non-negative integers, $ \ p_3(x) \ $ has only positive coefficients, so $ \ x \ = \ 1 \ $ cannot be one of its zeroes. Hence, $ \ p(x) \ $ has a factor limited to $ \ (x - 1)^3 \ \ . $