The gigantic polynomial

By the Remainder Factor Theorem , we know that, for a polynomial $P(x)$ , we have,

$(ax-b)\mid P(x)\iff P\left(\frac{b}{a}\right)=0~,~a\neq 0~~\ldots~(i)$

Now, consider the given polynomial as $P(x)$ and calculate $P(-1),P(0),P(1)$ . We see that only $P(-1)=0$ and $P(0),P(1)\neq 0$ . From these values and $(i)$ , we have,

$P(0)\neq 0\implies ax\not\mid P(x)~,~a\neq 0~~\ldots~(ii)$

$P(1)\neq 0\implies (x-1)\not\mid P(x)~~\ldots~(iii)$

$P(-1)=0\implies (x+1)\mid P(x)~~\ldots~(iv)$

By $(ii)$ , we can rule out options $x$ and $\dfrac{x}{2}$ . By $(iii)$ , we can rule out the option $(x-1)$

By $(iv)$ , we can conclude that $(x+1)$ divides $P(x)$ .

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Note: $x\mid y$ means that $y$ is divisible by $x$ .

Let

$P(x)=x^{9999}+x^{8888}+x^{7777}+\dots+x^{1111}+1$

From the Remainder Factor Theorem , $(x-a)$ is a factor of $P(x)$ if and only if $P(a)=0$ .

Now note that,

$P(-1)=(-1)^{9999}+(-1)^{8888}+(-1)^{7777}+\dots+(-1)^{1111}+1=0$

Therefore, $(x-(-1))=(x+1)$ is a factor of $P(x)$ and we are done.