Question

Phân tích thành nhân tử :

\(x^4+2004x^2+2003x+2004\)

Answer 1

       \(x^4+2004x^2+2003x+2004\)

\(=x^4-x+2004x^2+2004x+2004\)

\(=x\left(x^3-1\right)+2004\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2004\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2004\right)\)

Answer 2

\(x^4+2004x^2+2003x+2004\)

\(=x^4+2004x^2+2004x-x+2004\)

\(=\left(x^4-x\right)+2004\left(x^2+x+1\right)\)

\(=x\left(x^3-1\right)+2004\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2004\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2004\right)\)