How to factorize x ^ 5 + X + 1 = 0

How to factorize x ^ 5 + X + 1 = 0

x^5+x+1=x^5-x^2+x^2+x+1=x^2(x^3-1)+x^2+x+1=[x^2(x-1)+1](x^2+x+1)=(x^3-x^2+1)(x^2+x+1)

Decompose the factor and write the result 8A (x-a) + 4B (A-X) - 6C (x-a)=______ ．

8A (x-a) + 4B (A-X) - 6C (x-a) = 2 (x-a) (4a-2b-3c)

If PA Pb, then Po bisects APB. Why

(1) (2) the question is to make a vertical line & lt; P1, make OC perpendicular to AP, OD perpendicular to BP, and use the distance between the centers of equal chords to be equal, which means OC = OD, so Po bisector angle APB. (the points with equal distance to both sides of the angle are on the bisector line of this angle) 2 is the same, make a vertical line 3, connect PA and Pb, which can prove that Pao is equal to PbO by SSS, and then the angles are equal & lt; P

As shown in the picture, there is a river between villages a and B. now we need to build a bridge on the river which is perpendicular to the river bank and make villages a and B closest to each other. Try to determine the location of the bridge and explain the reasons
It's a theorem of translation

Use a line segment to connect two points a and B, find the intersection of this line segment and the middle line of the river, and build the bridge with this intersection as the midpoint of the bridge, so that the two villages can be closest
Geometry principle: the shortest line between two points

As shown in the figure, it is known that OC and od are two rays in ∠ AOB, OE bisects ∠ AOC, of bisects ∠ BOD. (1) if ∠ AOB = 132 °, COD = 22 °, calculate the degree of ∠ EOF; (2) if ∠ EOF = α, ∠ cod = β, calculate the degree of ∠ AOB. (expressed by algebraic formula containing α and β)

(1) According to the meaning of the subject, we get ∠ AOB = ∠ EOF + ∠ aoe +∠BOF=α+α-β=2α-β．

What is the necessary and sufficient condition for the quadratic equation x ^ 2 + ax + B = 0 to have two positive real roots?
A necessary and sufficient condition for the quadratic equation x ^ 2 + ax + B = 0 to have two positive real roots

Firstly, Δ ≥ 0 means a-4b ≥ 0, secondly, x1 × x2 > 0, X1 + x2 > 0 means - a > 0 (= >) deduces A0

Given the circle M: (x + 1) * 2 + y * 2 = 4, a (- 2,0), B (2,0), the moving point P in the circle m satisfies PA × Pb = Po * 2, the value range of point multiplication of vector pa by vector Pb is obtained
The answer is [- 2,6),

I suspect there is something wrong with this question. The reasons are as follows
Let P (x, y)
PA=(-2-x,-y)
PB=(2-x,-y)
PA*PB=（-2-x)(2-x)+y^2=x^2-4+y^2
PO^2=x^2+y^2
x^2-4+y^2=x^2+y^2
Can this be equal?

Solving 4 inequalities: 3x2 + 2x-1 ≥ 0 x2-3x-10 > 0 4x2-11x-3 < 0 - x2 + 3x + 10 > 0

Inequality one 3x2 + 2x-1 ≥ 0 (3x-1) (x + 1) ≥ 0 (3x-1) ≥ 0 (x + 1) ≥ 0 (x + 1) ≥ 0 x ≥ 1 / 3 and X ≥ - 1, so the solution satisfying the condition is x ≥ 1 / 3 inequality two x2-3x-10 > 0 (X-5) (x + 2) > 0 (X-5) > 0 (x + 2) > 0 x > 5 and X ＞ - 2, so the solution satisfying the condition is x > 5 inequality three 4x2-11x-3 < 0

In the known parallelepiped abcd-a1b1c1d1, the bottom surface is a square and the side edge Aa1 is B,

The square of x-7x-30

Cross multiplication:
-10 + 3=-7
-10*3=-30
So the factorization is as follows
（x-10）（x+3）