How do you factorize and find roots of the polynomial (1+x)n−xn (polynomials, complex numbers, roots, irreducible polynomials, weierstrass factorization, math)?

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How do you factorize and find roots of the polynomial  (1+x)n−xn  (polynomials, complex numbers, roots, irreducible polynomials, weierstrass factorization, math)?

nick1337 · Accepted Answer

Step 1 You may write that equation as follows (1+1x)n=1 from which you deduce that 1+1x should be a root of unity of order n in the complex plane, excluding the trivial root which is 1 itself. Hence we get n−1 solutions xk=1exp⁡(i2kπn)−1k=1.2⋯,n−1 Notice that for every even power n and k=n2 you’ll have xn2=−12 and in all the other cases the roots of this equation are xk=exp⁡(−i2kπn)−12(1−cos⁡(2kπn))=cos⁡(cos⁡(2kπn)−1−isin⁡(2kπn)2(1−cos⁡(2kπn))=−12−isin⁡(2kπn)4sin2⁡(kπn)=−12(1+cot⁡(kπn)) Once you have all the n−1 roots, the factorization of that polynomial of degree n−1 is trivial.

star233 · Accepted Answer

Step 1 Let: P(x)=(1+x)n−xn and ω=e2iπn;n∈N⋆ We have: d∘(P)=n−1 Case 1: n=0 If n=0 then ∀x∈C:P=0 Case 2 n=1 If n=1 then ∀x∈C:P=1 Case 3: n≥2 Now we assume: n≥2 P(0)≠0 so x≠0 P(x)=0⇔(1+x)n=xn⇔(x+1x)n=1⇔x+1x=ωk  with k∈1;2;⋯;n−1 ⇔x=1ωk−1=1e2ikπn−1 with k∈1;2;;⋯;n−1 ⇔x=1eikπn(eikπn−e−ikπn) with k∈1;2;⋯;n−1 ⇔x=eikπn2isin⁡(ikπn)=cos⁡(ikπn)−isin⁡(ikπn)2isin⁡(ikπn) with k∈1;2;⋯;n−1 ⇔x=xk=−12−i12cot⁡(kπn) with k∈1;2;⋯;n−1 So: ∀x∈C,∀n∈N,n≥2:P(x)=∏k=1n−1(x−xk)=∏k=1n−1(x+12+i12cot⁡(kπn))

How do you factorize and find roots of the polynomial (1+x)^{n}-x^{n} (polynomials,

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