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:
\(=t\left(t+2\right)-24=t^2+2t-24\)
\(=\left(t-4\right)\left(t+6\right)\)\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
