Factor x^105 + 1.

Step-by-step solution

1. 1

   Check the parity of the exponent first. A sum $x^{n}+1$ factors over the integers with $(x+1)$ as a factor exactly when $n$ is odd. Test it directly with the factor theorem: substituting $x=-1$ gives

   $(-1)^{105}+1=-1+1=0\ \checkmark$

   since $105$ is odd. So $x+1$ divides $x^{105}+1$ with no remainder.

2. 2

   Apply the sum-of-odd-powers identity. With $a=x$, $b=1$ and $n=105$:

   $a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2}b+a^{n-3}b^{2}-\cdots-ab^{n-2}+b^{n-1}\right)$

   Every power of $b=1$ equals $1$, so the cofactor collapses to alternating powers of $x$ alone.

3. 3

   Write the factorisation.

   $x^{105}+1=(x+1)\left(x^{104}-x^{103}+x^{102}-\cdots-x+1\right)$

   The cofactor has $105$ terms of degrees $104$ down to $0$, with signs alternating. Since the top degree $104$ is even, the leading sign is $+$ and — with an odd number of terms — the constant term is $+1$ as well.

4. 4

   Verify numerically. Evaluating both sides in exact integer arithmetic at $x=2,3,5,-2,-3$ gives equality every time ✓. Degrees also check: $1+104=105$ ✓.

5. 5

   Note that this is not the complete factorisation. Since $105=3\times5\times7$, further splitting is possible along every divisor. For instance $x^{105}+1=\left(x^{35}\right)^{3}+1$ also factors as $\left(x^{35}+1\right)\left(x^{70}-x^{35}+1\right)$. Fully factored over $\mathbb{Q}$, $x^{105}+1$ breaks into cyclotomic polynomials $\Phi_{d}(x)$ for the divisors $d$ of $210$ that do not divide $105$ — the $(x+1)$ form above is just the first, most useful split.