How do you find the remainder when dividing polynomials?

Question

Suppose that $f=(x+1)^{100}+(x-3)^{100}$ and $g=x^2-2x-3$. What is the remainder of $\frac{f}{g}$? I know there must be something connected to the roots of $g$, but how I can use that? Also I know that the remainder's degree is less or equal to 1. I don't want the response, I want to find that by myself, hints are welcomed.

Answer 1

If you put $$f(x)=q(x)g(x)+r(x)$$ where $r(x)$ has degree $1$, and you are not interested at all in $q(x)$, what can you do to make life easier?

Answer 2

Hint

We have that $$f(x)=(x^2-2x-3)q(x)+ax+b.$$ Since $$x^2-2x-3=(x-3)(x+1)$$ we get

$$f(-1)=-a+b$$ and $$f(3)=3a+b.$$

Answer 3

Complicated solution with binomial theorem:

If you write $x-1=t$ then $q(x) = t^2-4$ and $$p(x) = 2\Big[{100\choose 0}t^{100}+...+ 2^{96}{100\choose 96}t^4 +2^{98}{100\choose 98}t^2 + 2^{100}\Big]$$

Now let $s=t^2$ then $q(x) = s-4= q_1(s)$ and

$$p(x) = 2\Big[{100\choose 0}s^{50}+...+ 2^{96}{100\choose 96}s^2 +2^{98}{100\choose 98}s + 2^{100}\Big]= p_1(s)$$

So the remanider when $p_1(s)$ is divided by $s-4$ is $$p_1(4)=2\Big[{100\choose 0}2^{100}+...+ 2^{100}{100\choose 96} +2^{100}{100\choose 98} + 2^{100}\Big]$$ $$ = 2^{101}\Big[{100\choose 0}+...+ {100\choose 96} +{100\choose 98} + 1\Big] $$ $$ = 2^{100}\Big((1+1)^{100}+(1-1)^{100}\Big) =\boxed{2^{200}}$$

Answer 4

Recall that $$\newcommand{\rem}{\operatorname{rem}} \rem\!\big(p(x)s(x),q(x)s(x)\big)=s(x)\rem\!\big(p(x),q(x)\big) $$ and that $$ \rem\!\big(p(x),x-a\big)=p(a) $$ Now, since $x^2-2x-3=(x-3)(x+1)$, what we are looking for is $$ \begin{align} &\rem\left((x+1)^{100},(x-3)(x+1)\right)+\rem\left((x-3)^{100},(x-3)(x+1)\right)\\ %&=(x+1)\rem\left((x+1)^{99},x-3\right)+(x-3)\rem\left((x-3)^{99},x+1\right)\\ %&=(x+1)(3+1)^{99}+(x-3)(-1-3)^{99}\\[3pt] %&=4^{99}-3(-4)^{99}\\[3pt] %&=4^{100} \end{align} $$