Question

Phân tích thành nhân tử :

a ) \(3x^3-7x^2+17x-5\)

b ) \(x^4+2011x^2+2010x+2011\)

Answer 1

a ) \(3x^3-7x^2+17x-5\)

\(=\left(3x^2-x^2\right)-\left(6x^2-2x\right)+\left(15x-5\right)\)

\(=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)

\(=\left(x^2-2x+5\right)\left(3x-1\right)\)

b \(x^4+2011x^2+2010x+2011\)

\(=x^4-x+2011x^2+2011x+2011\)

\(=x\left(x^3-1\right)+2011\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2011\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2011\right)\)