How to find the remainder when a large degree polynomial is divided by a cubic polynomial with repeated roots.

Question

I was solving a question in which we had to find, as a step, the remainder when $x^{32}$ was divided by $x^3-x^2-x+1$. at first I tried to find it by using the remainder theorem in parts, once for each factor and then combine $3$ equations to find the remainder. But the divisor has a repeated root as it's factorisation is $(x-1)^2(x+1)$. so the system of equation turned out to have infinite solutions (as there were $3$ unknowns and $2$ equations) of the form $\alpha x^2 + \beta$ such that $\alpha + \beta = 1$. through matlab it can be confirmed that the remainder is $16x^2 -15$, but I could not derive it by hand. how can we do so?

Answer 1

You need one more equation involving the derivative at $x = 1$. It's a bit tedious to write the general statement but here is how it goes in this case. Write

$$x^{32} = (x - 1)^2 (x + 1) q(x) + r(x)$$

where $q$ is the quotient and $r$ is the remainder, so that $\deg r \le 2$. Substituting $x = \pm 1$ gives $r(1) = r(-1) = 1$, but this is only two equations. To get our third equation we need to differentiate once, getting

$$32x^{31} = (x - 1)^2 \left( \frac{d}{dx} (x+1)q(x) \right) + 2(x-1)(x+1)q(x) + r'(x)$$

and the key point is that the middle bit is still divisible by $x - 1$, so substituting $x = 1$ gives $r'(1) = 32$. This gives three equations for the coefficients of $r$ which can now be solved. Explicitly, $r(1) = r(-1) = 1$ gives that $r(x) = a(x + 1)(x - 1) + 1$ for some constant $a$, and $r'(1) = 32$ gives $a = 16$ as desired.

In general if the polynomial you're dividing by has a root of order $r$ then you need to differentiate at that root $r-1$ times.

Answer 2

We can also use (Easy) CRT. Let $\, t = x\!-\!1,\ $ so $\ (x\!-\!1)^2(x+1) = t^2(t+2)$

$\quad\ \ f = (1\!+\!t)^{2n}\equiv (-1)^{2n}\equiv 1\ \pmod{\!t\!+\!2},\ $ by $\ f(t)\bmod \color{#c00}{t\!+\!2\, = f(-2)}$

$\quad\ \ f = (1\!+\!t)^{2n}\equiv \color{#0a0}{1+2nt}\qquad\ \pmod{\!\color{#0a0}{t^2}},\quad\ $ by $ $ Binomial Theorem

$\!\!\begin{align}\overset{\small\rm\color{#0af}{CRT}}\Rightarrow f \equiv \color{#0a0}{1\!+\!2nt}+\color{#0a0}{t^2}\left[\dfrac{-2nt}{t^2}\smash[b]{\underbrace{\bmod \color{#c00}{t\!+\!2}}_{\color{#c00}{\large t\ \equiv\ -2}}}\right]&\equiv\, 1\!+\!2nt+t^2[\color{#c00}4n/\color{#c00}4]\pmod{t^2(t\!+\!2)}\\ &\equiv\, n(\underbrace{1\!+\!t}_{\large x})^2\!+1\!-\!n,\quad\ {\rm OP\ is}\ \,n\!=\!16\end{align}$

Answer 3

Let $P(x)=(x-1)^2(x+1)$ . Then, note that, $P'(x)=(x-1)(3x+1)$ .

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You have :

$$ \small{\begin{align}x^{32}=P(x)Q_1(x)+ax^2+bx+c \end{align}} $$

Setting $x=±1$, yields :

$$ \begin{align}&\begin{cases}a+b+c=1\\ a-b+c=1\end{cases}\\ \implies &2b=1-1=0\\ \implies &b=0\end{align}$$

Then, taking derivative leads to :

$$32x^{31}=(x-1)Q_2(x)+2ax$$

Again setting $x=1$, we conclude :

$$ \begin{align}&2a=32\\ \implies &\begin{cases}a=16\\ c=1-16=-15\end{cases}\end{align} $$

Thus, $R(x)=16x^2-15$ .

Answer 4

The remainder of the division of $(y+1)^n$ by $y^2$ is $n y + 1$. Now, take $y = x^k-1$. Conclude:

The remainder of the division of $x^{k n}$ by $(x^k-1)^2$ is $n x^k -(n-1)$. So it will be under division by $D(x)$, a divisor of $(x^k-1)^2$ of degree $d> k$.