Question

在实数范围内分解因式(x2+x-6)(x2+x+1)+12

Answer 1

(x2+x-6)(x2+x+1)+12
=（x^2+x)^2-5(x^2+x)-6+12
=(x^2+x)^2-5(x^2+x)+6
=(x^2+x-2)(x^2+x-3)
=(x-1)(x+2)(x^2+x-3)

Answer 2

原式=(x²+x)²-5(x²+x)-6+12
=(x²+x)²-5(x²+x)+6
=(x²+x-2)(x²+x-3)
=(x+2)(x-1)[(x+1/2)²-13/4]
=(x+2)(x-1)(x+1/2+√13/2)(x+1/2-√13/2)