# Problem Of The Week #379 Aug 15th, 2019

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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Find (without calculus) a fifth degree polynomial $P(x)$ such that $P(x)+1$ is divisible by $(x-1)^3$ and $P(x)-1$ is divisible by $(x+1)^3$.

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#### anemone

##### MHB POTW Director
Staff member
Congratulations to the following members for their correct solution!

1. Olinguito
2. castor28
3. Opalg
4. MegaMoh

Solution from Olinguito :
Let
$$P(x)\ \equiv\ (ax^2+bx+c)(x-1)^3-1\ \equiv\ (dx^2+ex+f)(x+1)^3+1.$$
By comparing the coefficients of $x^5$, we immediately have $d=a$.

Comparing the constant terms gives $f+1=-c-1$ $\implies$ $f=-c-2$.

And comparing the coefficients of $x^4$ gives $-3a+b=3d+e=3a+e$ $\implies$ $e=-6a+b$. Thus:
$$(ax^2+bx+c)(x-1)^3-1\ \equiv\ (ax^2+[-6a+b]x-c-2)(x+1)^3.$$
The coefficient of $x^2$ on the LHS is $-a+3b-3c$; on the RHS it is $a+3(-6a+b)+3(-c-2)=-17a+3b-3c-6$. Therefore
$$-a+3b-3c\ =\ -17a+3b-3c-6$$
$\implies$ $16a=-6$ $\implies$ $a=-\dfrac38$.

The coefficient of $x^4$ on the LHS is $3a-3b+c$; on the RHS it is $3a+3(-6a+b)-c-2$. So
$$-3b+c\ =\ 3(-6a+b)-c-2=-18a+3b-c-2=\frac{27}4+3b-c-2=\frac{19}4+3b-c$$
$\implies$ $-24b+8c\ =\ 19\ \ldots\ \fbox1$.

And finally $P(-1)=1$ implies
$$\left(-\frac38-b+c\right)(-1-1)^3-1=1\ \implies\ 8b-8c=-1\ \ldots\ \fbox2.$$
Solving the simultaneous equations $\fbox1$ and $\fbox2$ gives $b=-\dfrac98$ and $c=-1$.

Hence
$$P(x)\ =\ \left(-\frac38x^2-\frac98x-1\right)(x-1)^3-1\ =\ \boxed{-\frac38x^5+\frac54x^3-\frac{15}8x}.$$

Alternate solution from castor28 :
We know that:
\begin{align*}
P(x) &\equiv +1 \pmod{(x+1)^3}\\
P(x) &\equiv -1 \pmod{(x-1)^3}
\end{align*}
and we must find $P(x)$.

We use the Chinese Remainder Theorem (over the ring $\mathbb{Q}[x]$). As usual, the first step is to find polynomials $u(x)$ and $v(x)$ that satisfy Bézout's identity:
$$u(x)(x+1)^3 + v(x)(x-1)^3 = \gcd((x+1)^3,(x-1)^3) = 1$$
Since $\mathbb{Q}[x]$ is a Euclidean domain, we can use the Extended Euclidean algorithm. The computation proceeds as follows:
$$\begin{array}{c|c|c|c} q_n&r_n&u_n & v_n\\ \hline & x^3 + 3x^2 + 3x + 1 & 1 & 0\\ & x^3 - 3x^2 + 3x - 1 & 0 & 1\\ 1 & 6x^2 + 2 & 1 & -1\\ \dfrac{x}{6}-\dfrac12 & \dfrac{8x}{3} & -\dfrac{x}{6}+\dfrac12 & \dfrac{x}{6}+\dfrac12\\ \dfrac{9x}{4} & 2 & \dfrac{3x^2}{8} - \dfrac{9x}{8} + 1 & -\dfrac{3x^2}{8} - \dfrac{9x}{8} - 1 \end{array}$$
In this table, $q_n$ is the polynomial quotient $r_{n-2}/r_{n-1}$, $r_n$ is the remainder, and line $L_n$ is computed as $L_{n-2}-q_n L_{n-1}$. In each line, we have $r_n=u_n(x+1)^3 + v_n(x-1)^3$ (by construction for the first two lines and by induction for the other lines).

This gives us:
\begin{align*}
u(x) &= \dfrac{3x^2}{16} - \dfrac{9x}{16} + \dfrac12\\
v(x) &= -\dfrac{3x^2}{16} - \dfrac{9x}{16} - \dfrac12
\end{align*}

and we get:
\begin{align*}
P(x) &= (+1)v(x)(x-1)^3 + (-1)u(x)(x+1)^3\\
&= -\dfrac{3x^5}{8} + \dfrac{5x^3}{4} - \dfrac{15x}{8}
\end{align*}
Since this is defined modulo $(x+1)^3(x-1)^3$ and $P(x)$ must have degree $5$, the solution is unique.

Alternate solution from Opalg :
Geometrically, the given information says that the graph of $P(x)$ has cubic points of inflection at the points $(1,-1)$ and $(-1,1)$. So the graph ought to look something like this:

This makes it look as though the graph should be symmetric under a $180^\circ$ rotation around the origin. In that case, $P(x)$ will be an odd function. Since it is a polynomial, that means that it will have no terms of even degree.

If $P(x)$ is a fifth degree polynomial and $P(x) + 1$ is divisible by $(x-1)^3$ then $P(x) = (x-1)^3(ax^2 + bx + c) -1$ for some constants $a,b,c$. Then \begin{aligned}P(x) &= (x^3 - 3x^2 + 3x - 1)(ax^2 + bx + c) - 1 \\ &= ax^5 + (b-3a)x^4 + (c-3b+3a)x^3 + (-3c+3b-a)x^2 + (3c-b)x + (-c - 1).\end{aligned} If the coefficients of all the even powers of $x$ are zero then $$b - 3a = -3c + 3b - a = -c-1 = 0.$$ From that, it easily follows that $a = -\tfrac38$, $b = -\tfrac98$, $c = -1$. Substituting those values into the formula for $P(x)$ gives the coefficient of $x^3$ as $\tfrac54$ and the coefficient of $x$ as $-\tfrac{15}8$. Therefore $\boxed{P(x) = -\tfrac38x^5 + \tfrac54x^3 - \tfrac{15}8x}$.

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