Question

Phân tích đa thức thành nhân tử :

a, \(x^8+x^4+1\)

b, \(x^4+2008x^2+2007x+2008\)

Answer 1

\(x^8+x^4+1\)

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

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

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

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

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

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

\(x^4+2008x^2+2007x+2008\)

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

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

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

\(=\left(x^2+x+1\right)\left(x+2007\right)\)