# Prove that the polynomial x^9999+x^8888+x^7777+...+x^1111+1 is divisible by x+1

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Let's see something before going to your actual question.

Number of terms in above polynomial = 10

If we exclude +1 , We have 9 terms.

We know that,

-1 ^ odd = -1

-1 ^ even = 1

Now,

Let's consider that

f(x) = x^9999+x^8888+x^7777+...+x^1111+1 .

If f(x) is divisible by ( x + 1 ) then f(-1) = 0 .

We will have to prove that, f(-1) = 0

f(x) = x^9999 + x^8888 + ......... + x^2222 + x^1111 + 1

f(-1) = (-1)^9999 + (-1)^8888 +......+ (-1)^2222 + (-1)^ 1111 + 1

f(-1) = -1 + 1 + ... + 1 - 1 + 1

f(-1) = 0

Therefore, x^9999+x^8888+x^7777+...+x^1111+1 is divisible by x+1 .

Number of terms in above polynomial = 10

If we exclude +1 , We have 9 terms.

We know that,

-1 ^ odd = -1

-1 ^ even = 1

Now,

Let's consider that

f(x) = x^9999+x^8888+x^7777+...+x^1111+1 .

If f(x) is divisible by ( x + 1 ) then f(-1) = 0 .

We will have to prove that, f(-1) = 0

f(x) = x^9999 + x^8888 + ......... + x^2222 + x^1111 + 1

f(-1) = (-1)^9999 + (-1)^8888 +......+ (-1)^2222 + (-1)^ 1111 + 1

f(-1) = -1 + 1 + ... + 1 - 1 + 1

f(-1) = 0

Therefore, x^9999+x^8888+x^7777+...+x^1111+1 is divisible by x+1 .

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