Question

Phân tích thành nhân tử bằng cách đặt biến phụ

a,[x^2+x]^2+4x^2+4x-12

b,[x^2+4x+8]^2+3x^3+14x^2+24x

Answer 1

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

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

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

b: \(\left(x^2+4x+8\right)^2+3x^3+14x^2+24x\)

\(=\left(x^2+4x+8\right)^2+3x^3+12x^2+24x+2x^2\)

\(=\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)