Question

Phân tích thành nhân tử:

a) (x²+8x+7)(x+3)(x+5) + 15

b) (x+2)(x+3)(x+4)(x+5) - 24

Answer 1

a)\(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(t=x^2+8x+7\) ta có:

\(=t\left(t+8\right)+15=t^2+8t+15\)

\(=\left(t+3\right)\left(t+5\right)\)\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)

\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)

b)\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(t=x^2+7x+10\) ta có:

\(=t\left(t+2\right)-24=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)