Question

Phân tính đa thức thành nhân tử :

a) x⁵ + x⁴ + 1

b) x⁸ + x +1

c) x⁸ + x⁴ + 1

Answer 1

Answer 2

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

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

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

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

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

\(=\left(x^2+x+1\right)\left(x^4-x+1\right)\)

b) \(x^8+x+1=\left(x^8-x^2\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x^6-1\right)+\left(x^2+x+1\right)\)

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

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

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

c) \(x^8+x^4+1=\left(x^8-x^2\right)+\left(x^4-x\right)+\left(x^2+x+1\right)\)

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

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

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

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

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)