Question

Phân tích đa thức thành nhân tử ( PP hệ số bất định ):

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

Đặt biến phụ : \(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)-18\)

\(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

Answer 1

Bài 2:

a: \(=\left(4x^2+6x+2x+3\right)\left(x^2+2x+1\right)-18\)

\(=\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18\)

Đặt x^2+2x=a

\(A=\left(4a+3\right)\left(a+1\right)-18\)

\(=4a^2+7a-15\)

\(=4a^2+12a-5a-15=\left(a+3\right)\left(4a-5\right)\)

\(=\left(x^2+2x+3\right)\left(4x^2+8x-5\right)\)

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

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

b: \(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)

\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)