Phân tích đa thức sau thành nhân tử:

- Hướng dẫn giải

$\begin{array}{l}\left(x^2+2x\right)\left(x^2+4x+3\right)-24\\ =\left(x^2+2x\right)\left(x^2+4x+3\right)-24\\ =x\left(x+2\right)\left[x\left(x+1\right)+3\left(x+1\right)\right]-24\\ =x\left(x+2\right)\left(x+1\right)\left(x+3\right)-24\\ =\left[x.\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]-24\\ =\left(x^2+3x\right)\left(x^2+3x+2\right)-24\\ ={\left(x^2+3x\right)}^2+2\left(x^2+3x\right)-24\\ ={\left(x^2+3x\right)}^2+6\left(x^2+3x\right)-4\left(x^2+3x\right)-24\\ =\left[{\left(x^2+3x\right)}^2+6\left(x^2+3x\right)\right]-\left[4\left(x^2+3x\right)+24\right]\\ =\left(x^2+3x\right)\left(x^2+3x+6\right)-4\left(x^2+3x+6\right)\\ =\left(x^2+3x-4\right)\left(x^2+3x+6\right)\end{array}$

Vậy $\left(x^2+2x\right)\left(x^2+4x+3\right)-24=\left(x^2+3x-4\right)\left(x^2+3x+6\right)$