Question

Phân tích đa thức thành nhân tử: \(\left(x^2+x\right)^2+4x^2+4x-12\)

Answer 1

\(A=\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left[\left(x^2+x\right)+2\right]^2-4^2\)\(A=\left[\left(x^2+x\right)+2-4\right]\left[\left(x^2+x\right)+2+4\right]\)

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

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

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

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