Question

phân tích đa thức thành nhân tử :

(x+2)(x+3)(x+4)(x+5)-8

Answer 1

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

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

\(=\)\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt \(x^2+7x+11=t\) ta có : 

\(=\)\(\left(t-1\right)\left(t+1\right)-8\)

\(=\)\(t^2-1-8\)

\(=\)\(t^2-9\)

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

\(=\)\(\left(x^2+7x+11-3\right)\left(x^2+7x+11+3\right)\)

\(=\)\(\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

Chúc bạn học tốt ~