a) \(=\left[\left(x^2+x\right)^2-2.7\left(x^2+x\right)+7^2\right]-25=\left(x^2+x-7\right)^2-5^2=\left(x^2+x-7-5\right)\left(x^2+x-7+5\right)=\left(x^2+x-12\right)\left(x^2+x-2\right)=\left[\left(x^2+4x\right)-\left(3x+12\right)\right]\left[\left(x^2+2x\right)-\left(x+2\right)\right]=\left[x\left(x+4\right)-3\left(x+4\right)\right]\left[x\left(x+2\right)-\left(x+2\right)\right]=\left(x-3\right)\left(x+4\right)\left(x-1\right)\left(x+2\right)\)
c) \(=\left(x^4-x^3\right)+\left(3x^3-3x^2\right)+\left(8x^2-8x\right)+\left(12x-12\right)=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=\left(x^3+3x^2+8x+12\right)\left(x-1\right)=\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\left(x-1\right)=\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)
e) \(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=t\)
\(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15=\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left[\left(x^2+6x\right)+\left(2x+12\right)\right]=\left(x^2+8x+10\right)\left[x\left(x+6\right)+2\left(x+6\right)\right]=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)
