Dividing a Huge Polynomial

By remainder factor theorem , if $P(x)$ is divisible by $x-c$ , then $P(c) = 0$ . In this case, if $P(x) = x^{9999} + x^{8888} + x^{7777 }+\cdots+ x^{1111} + 1$ is divisible by $(x+1)$ , then $P(-1) = 0$ .

Now we want to check whether $P(-1) = 0$ is true or not, if it is indeed true then the polynomial in question is divisible by $(x+1)$ ; otherwise it is not divisible by $(x+1)$ .

Recall that
$(\text{negative number})^\text{odd number}$ yields a negative number; while
$(\text{negative number})^\text{even number}$ yields a positive number.

Substituting $x = -1$ gives,

$P(-1) = (-1)^{9999} + (-1)^{8888} + (-1)^{7777} + \cdots + (-1)^{1111} + 1$

Computing the powers of each of these numbers gives

$\begin{aligned} (-1)^{9999} &=& (-1)^{7777} = (-1)^{5555}= (-1)^{3333}= (-1)^{1111} = -1 \\ (-1)^{8888} &=& (-1)^{6666} = (-1)^{4444}= (-1)^{2222}= 1 \end{aligned}$

Thus, we can evaluate $P(-1)$ with ease!

$P(-1) = \cancelto{0}{ (-1) + (1) } + \cancelto{0}{ (-1) + (1) } + \cancelto{0}{ (-1) + (1) } + \cancelto{0}{ (-1) + (1) } \cancelto{0}{ (-1) + (1) } = 0$

And because we have shown that $P(-1) = 0$ is indeed true, so the polynomial in question is divisible by $(x+1)$ .

[x^(2y+1) +1 ] / [x+1] = -1 + x - x^2 +x^3 ... - x^2y

[x^(2m) - 1] / [x+1] = 1 -x +x^2 -x^3 ... -x^(2m-1)

x^9999 + x^8888 + x^7777 + x^6666 ... + x+1 = [x^9999 +1] + [x^8888 -1] +[x^7777+1] +[x^6666 -1]... + x+1