Polynomial problem with two conditions

Question

I have to find $P(0)$ from the polynomial with minimum degree given that $$(x-1)^3|(P(x)+1)$$ $$(x+1)^3|(P(x)-1)$$ Plugging in $x=\pm 1$ gets something nice, also division by a polynomial of third order gives successively: $$P(1)+1 =0; \ P'(1)=0; \ P''(1)=0$$ $$P(-1)-1 =0; \ P'(-1)=0; \ P''(-1)=0$$

Furthermore since $P(1)=-1$ and $P(-1)=1$, also $P(1)=-P(-1)$ and $$P'(1)=P''(1)=P'(-1)=P''(-1)$$ I don't see how to use this stuff.

Answer 1

Suppose $P\in\mathbb{Q}[x]$ is a polynomial of least degree such that \begin{align*} (x+1)^3&{\,\mid\,}(P(x)-1)\\[4pt] (x-1)^3&{\,\mid\,}(P(x)+1)\\[4pt] \end{align*} Equivalently, $P\in\mathbb{Q}[x]$ is a polynomial of least degree such that \begin{align*} P(x)&=A(x)(x+1)^3+1\\[4pt] P(x)&=B(x)(x-1)^3-1\\[4pt] \end{align*} for some polynomials $A,B\in\mathbb{Q}[x]$.

Necessarily, $A,B$ have the same degree.

Let $u=(x+1)^3$ and let $v=(x-1)^3$.

If $A,B\in\mathbb{Q}[x]$ are polynomials of least degree satisfying $Au-Bv=-2$, then we have $Au+1=Bv-1$, hence $P=Au+1$ is a polynomial in $\mathbb{Q}[x]$ of least degree satisfying the specified requirements.

The equation $Au-Bv=-2$ is equivalent to $-\bigl({\large{\frac{A}{2}}}\bigr)u+\bigl({\large{\frac{B}{2}}}\bigr)v=1$, hence we have $A=-2a$ and $B=2b$, where $a,b\in\mathbb{Q}[x]$ are polynomials of least degree satisfying $au+bv=1$.

Since $u,v$ are relatively prime in $\mathbb{Q}[x]$, there are unique polynomials $a,b\in\mathbb{Q}[x]$ with

• $\deg(a) < \deg(v) = 3$$\\[4pt]$
• $\deg(b) < \deg(u) = 3$

such that $au+bv=1$, so these are the polynomials $a,b$ we want.

The polynomials $a,b$ can be found via the Extended Euclidean Algorithm.

Here are the steps . . . \begin{align*} (x+1)^3=(1)(x-1)^3+(6x^2+2)&\implies u-v=6x^2+2\\[4pt] 6(x-1)^3=(x-3)(6x^2+2)+16x&\implies (-x+3)u+(x+3)v=16x\\[4pt] 8(6x^2+2)=(3x)(16x)+16&\implies (3x^2-9x+8)u+(-3x^2-9x-8)v=16\\[4pt] \end{align*} from which we get \begin{align*} a&={\small{\frac{1}{16}}}(3x^2-9x+8)\\[4pt] b&={\small{\frac{1}{16}}}(-3x^2-9x-8)\\[4pt] \end{align*} which yields \begin{align*} A&=-2a=-{\small{\frac{1}{8}}}(3x^2-9x+8)\\[4pt] B&=2b={\small{\frac{1}{8}}}(-3x^2-9x-8)\\[4pt] \end{align*} hence \begin{align*} P&=Au+1\\[4pt] &=\left(-{\small{\frac{1}{8}}}(3x^2-9x+8)\right)(x+1)^3+1\\[4pt] &=-{\small{\frac{1}{8}}}(3x^5-10x^3+15x)\\[4pt] \end{align*}