Question

Phân tích đa thức thành nhân tử

a) \(x^5+x+1\)

b) \(x^5+x^4+1\)

c) \(x^8+x+1\)

d) \(x^8+x^7+1\)

Answer 1

\(d.x^8+x^7+1\\ =x^8+x^7+x^6-x^6+1\\ =x^6\left(x^2+x+1\right)-\left(x^6-1\right)\\ =x^6\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\\ =x^6\left(x^2+x+1\right)-\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left[x^6-\left(x-1\right)\left(x^3+1\right)\right]\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

Answer 2

Answer 3

\(a.x^5+x+1\\ =x^5-x^2+x^2+x+1\\ =x^2\left(x^3-1\right)+x^2+x+1\\ =x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left[x^2\left(x-1\right)+1\right]\left(x^2+x+1\right)\\ =\left(x^3-x^2+1\right)\left(x^2+x+1\right)\)

Answer 4

Bài 1 :

a, Ta có : \(x^5+x+1\)

= \(x^5+x^2-x^2+x+1\)

= \(x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

= \(x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

= \(\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)

b, Ta có : \(x^5+x^4+1\)

= \(x^5+x^4+x^3-x^3+1\)

= \(x^3\left(x^2+x+1\right)-\left(x^3-1\right)\)

= \(x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

= \(\left(x^2+x+1\right)\left(x^3-x+1\right)\)

Answer 5

\(b.x^5+x^4+1\\ =x^5+x^4+x^3-x^3+1\\ =x^3\left(x^2+x+1\right)-\left(x^3-1\right)\\ =x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =\left(x^3-x+1\right)\left(x^2+x+1\right)\)

Answer 6

\(c.x^8+x+1\\ =x^8+x^2-x^2+x+1\\ =x^2\left(x^6-1\right)+x^2+x+1\\ =x^2\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\\ =\left(x^5+x^2\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left[\left(x^5+x^2\right)\left(x-1\right)+1\right]\left(x^2+x+1\right)\\ =\left(x^6-x^5+x^3-x^2+1\right)\left(x^2+x+1\right)\)