Question

phân tích thành nhân tử:

a. \(\left(x^2+x\right)^2+4x^2+4x-12\) 

b. \(x^3+3x^2-4\)

Answer 1

b. \(x^3+3x^2-4\)  =\(x^3-x^2+4x^2-4\)

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

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

Answer 2

\(x^3+3x^2-4=x^3-1+3x^2-3\)

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

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

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

Answer 3

a) \(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=x^4+2x^3+x^2+4x^2+4x-12\)

\(=x^4+2x^3+5x^2+4x-12\)

\(=\left(x^4+2x^3\right)+\left(5x^2+10x\right)-\left(6x+12\right)\)

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

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

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

\(=\left(x+2\right)\left[x^2\left(x-1\right)+6\left(x-1\right)\right]\)

\(=\left(x+2\right)\left(x-1\right)\left(x^2+6\right)\) 

b) \(x^3+3x^2-4\)

\(=x^3-x^2+4x^2-4\)

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

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

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

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