Help with complex/polynomial question (1 Viewer)

[timmiitippii]

does anyone know how to find all the zeros of
P(x) = x^4 + x^3 + x^2 + x + 1

then hence factorise P(x) into irreducible factors over R

thank you in advance

 

[QuLiT]

i think you can do it like this

(x-1)(x^4 + x^3 + x^2+ x+1) using this, it's the roots of x^5-1=0 other than x=1.

so x= cis 2kpi/5 k= + or - 1 , + or - 2

then just multiply the conjugates in form (x-cis2pi/5)(x-cis-2pi/5) which is (x^2-x(2cos2pi/5) +1)(x^2 -x(2cos4pi/5) +1) <--- final factors over R

 

[Trebla]

P(x) = x⁴ + x³ + x² + x + 1 = (x⁵ - 1)/(x - 1) since the LHS is a geometric series
To find the zeros of P(x), we need find the zeroes of (x⁵ - 1) for x =/= 1 (i.e. non real solutions of x)
So solving x⁵ - 1 = 0 for x=/=1
x = cis 2π/5, cis 4π/5, cis 6π/5, cis 8π/5 (ignore cis 0 = 1 as we require non-real solutions)
= cis 2π/5, cis 4π/5, cis -2π/5, cis -4π/5 since co-efficients are real, we have complex conjugate roots
Hence P(x) = (x - cis 2π/5)(x - cis -2π/5)(x - cis 4π/5)(x - cis -4π/5)

However, we require the factorisation to be over the real field, so note that:
*cis θ + cis (-θ) = cos θ + isin θ + cos (-θ) + isin (-θ)
= cos θ + isin θ + cos θ - isin θ (as cos (-θ) = cos θ and sin (-θ) = -sin θ)
= 2cos θ
AND
*cis θ.cis (-θ) = cis 0 = 1
Thus,
P(x) = (x² - 2xcos 2π/5 + 1)(x² - 2xcos 4π/5 + 1)

 

[QuLiT]

.....
P(x) = (x² - 2cos 2π/5 + 1)(x² - 2cos 4π/5 + 1)

missing the x's

 

[timmiitippii]

thank you so much QuLiT and Trebla. Both of your explanations were very helpful