Question

Phân tích đa thưc thành nhân tử :

x⁴+2008x²+2007x+2008

Answer 1

\(x^4+2008x^2+2007x+2008\\ =x^4-x+2008\left(x^2+x+1\right)=x\left(x^3-1\right)+2008\left(x^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

Answer 2

Ta có: \(x^4+2008x^2+2007x+2008\)

\(=x^4-x+2008\left(x^2+x+1\right)\)

\(=x\left(x^3-1\right)+2008\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)