Question

Phân tích đa thức thành nhân tử

1, x⁵+x+1

2, x¹⁰+x⁵+1

Answer 1

\(\text{1. }x^5+x+1\\ \\=x^5+x+1+x^2-x^2\\ \\ =\left(x^5-x^2\right)+\left(x^2+x+1\right)\\ \\=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\\ \\ =x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ \\ =\left(x^3-x^2+1\right)\left(x^2+x+1\right)\)

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

Answer 2

1,x(x⁴+1)+1. 2.x⁵(x²+1)+1 Good luck to u