Question

Phân tích đa thức thành nhân tử

x¹⁰ + x⁸ + 1

Answer 1

\(x^{10}+x^8+1\)

\(=\left(x^{10}-x\right)+\left(x^8-x^2\right)+\left(x^2+x+1\right)\)

\(=x\left(x^9-1\right)+x^2\left(x^6-1\right)+\left(x^2+x+1\right)\)

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

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

Answer 2

\(x^{10}+x^8+1\\ =x^{10}+x^9+x^8-x^9-x^8-x^7++x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\\ =x^8\left(x^2+x+1\right)-x^7\left(x^2+x+1\right)+x^6\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+1\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^8-x^7+x^6-x^4+x^3-x+1\right)\)