Question

Phân tích đa thức thành nhân tử : 

(x+1)(x+5)(x+2)(x+4)-4 

Answer 1

\(\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-4\)

\(=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)

Đặt \(x^2+6x+5=t\)

\(\Rightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-4\)

\(=t\left(t+3\right)-4\)

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

\(=\left(t^2-t\right)+\left(4t-4\right)\)

\(=t.\left(t-1\right)+4\left(t-1\right)\)

\(=\left(t-1\right)\left(t+4\right)\)

\(=\left(x^2+6x+4\right)\left(x^2+6x+9\right)\)