Find the standard decomposition of F (x) = x ^ n + 1 in complex number field and real number field

Find the standard decomposition of F (x) = x ^ n + 1 in complex number field and real number field

x^n=-1=cosπ+isinπ
x=cos{[π+2(k-1)π]/n}+isin{[π+2(k-1)π]/n} (k=1,2,3,.,n)
  =cos[(2k-1)π/n]+isin[(2k-1)π/n] (k=1,2,3,.,n)
Let's send the pictures

How to factorize x ^ n + 1 in real and complex fields

Real range: when n is a multiple of 4, it can be decomposed; when n is a multiple of 2, it cannot be decomposed; when n is an odd number, it can be decomposed
X ^ n + 1 when n is odd
=(x+1)[x^(n-1)-x^(n-2)+x^(n-3)+…… -x+1]
When n is a multiple of 4, let n = 4m
x^n+1=x^4m+1=(x^2m+1)^2-2x^2m=(x^2m+1-√2x^m)(x^2m+1+√2x^m)
=(x^2m-√2x^m+1)(x^2m+√2x^m+1)
In the complex range, when n is odd
x^n+1
=(x+1)(x-x1)(x-x2)…… [x-x(n-1)]
among
x1=cos(π/n)+isin(π/n)
x2=cos(3π/n)+isin(3π/n)
……
x(n-1)=cos{[2(n-1)-1]π/n}+isin{[2(n-1)-1]π/n}
=cos[(2n-3)π/n]+isin[(2n-3)π/n]
In the range of complex number, when n is even
x^n+1
=(x-x1)(x-x2)…… (x-xn)
among
x1=cos(π/n)+isin(π/n)
x2=cos(3π/n)+isin(3π/n)
……
xn=cos[(2n-1)π/n]+isin[(2n-1)π/n]

Use 4 congruent right triangles to form an isosceles trapezoid. 2 kinds of spelling (quick, solve it today)
Describe the method, better with a picture
No isosceles. It's trapezoid

First use two of them to make a rectangle, and then use the long right angle side and the short right angle side as the top edge of the trapezoid

Objects with the same surface roughness and the mass ratio of M1 to m2 of 2:3 are placed on the horizontal table
Given that the horizontal thrust is f = 20n, it just moves in a straight line at a uniform speed, and then calculate the thrust on M1 and M2?

Uniform motion, so the resultant force is zero
F1=um1g F2=um2g
F1:F2=m1:m2=2:3
F1 + F2 = f F1 = 8N F2 = 12n

Integral operation: 1. (x + 2) (x + 3)

Square of X + 5x + 6

A bus runs along a straight road. When it starts from the station, the acceleration is 2 meters per second. It accelerates for 5 seconds, then runs at a constant speed for 2 minutes, and then stops
Turn off the engine, taxi 50 meters, and the speed is zero when you arrive at station B. ask for the average speed of the car from station a to station B. I'm not good at physics, please give me more advice,

Acceleration displacement: S1 = at & # 178 / 2 = 2x25 / 2 = 25m
Uniform velocity v = at = 10m / S
Uniform displacement: S2 = VT = 10x120 = 1200m
S3=50m t3=50/(10/2)=10s
Average velocity v '= S1 + S2 + S3 / (T1 + T2 + T3) = (1200 + 50 + 25) / (5 + 120 + 10) = 9.44m/s

The sum of the first n terms of the arithmetic sequence {an} is Sn, the square of (am-1) + (am + 1) - am = 0, s2m-1 = 38, what is m
M-1, M + 1, 2m-1, m are all footmarks

Very simple, square equals 0, not square or 0, let t = M-1, then at + at2-at1 = 0, A1 = 0

When the vehicle is in emergency braking, the speed is 12m / s, and the acceleration obtained by braking is 4m / s # 178;, the displacement within 1.5s after braking is calculated
(2) Time of 16m after braking (3) displacement within 5S after braking

V = V0 at; v = 12-1.5 * 4 = 6 not stopped
S=Vt-1/2at^2
s=12*1.5-0.5*4*1.5^2=13.5m
2) S = VO ^ 2 / (2a) = 18m not stopped
S=Vt-1/2at^2
16=12t-1/2at^2
t1=2,t2=4
t(max)=vo/a=12/4=3s
So when t = 2S, the displacement is 16m
3) Car movement 3S
So the displacement at 5S is s = s = 18m

Xiao Ming and Xiao Gang often practice running on the 400 meter long circular track. Xiao Ming's speed is 8 meters per second, and Xiao Gang's speed is 7 meters per second. If they are facing each other at the same time, they will be able to run
Xiao Ming and Xiao Gang often practice running on the 400 meter long circular track. Xiao Ming's speed is 8 m / s, Xiao Gang's speed is 7 m / s, and Xiao Gang's speed is 7 m / s. if they are facing each other at the same time, how many seconds will they meet for the first time
I know it's 89 seconds,

1. If a and 2b are reciprocal to each other - C and 2 / 2 D are opposite to each other
|X | = 3 find the value of X of 2ab-2c + D + 3
2. Using the number axis to find the minimum value of | X-2 | + | x-3 | + 1
3. One day, Xiao Ming was running on a north-south road. He started from a place and recorded his running situation every 10 minutes (south is the positive direction, unit: m): - 1008, 1100, - 976, 1010, - 827, 946. After one hour, he stopped to have a rest. At this time, what direction was he in a place? How far was he from a place? How many meters did Xiao Ming run?
okay,,,

A and 2b are reciprocal to each other - C and 2 / D are opposite to each other, so a * 2B = 1, - C + D / 2 = 02ab = 1, - 2C + D = 0.2ab-2c + D + 3 of x = 1 + 0 + X / 3 = x / 3 + 1, because | x | = 3, x = 3 or - 3, X / 3 + 1 = 2 or 02:2 zeros: x = 2, x = 3. When x is between these two points, | X-2 | + | x-3 | + 1 takes the minimum value = 1 + 1 = 23:1008 + 1100 + (...)