Question

因式分解 (x²+x+1)(x²+x+2)-12

分解因式

Answer 1

令x²+x = k

(x²+x+1)(x²+x+2)-12
= (k+1)(k+2) -12
= k²+3k +2 - 12
= k² +3k -10
= (k+5)(k-2)
= (x²+x+5)(x²+x-2)
= (x²+x+5)(x+2)(x-1)

Answer 2

(x²+x+1)(x²+x+2)-12
=[(x²+x)+1][(x²+x)+2]-12
=(x²+x)²+3(x²+x)+2-12
=(x²+x)²+3(x²+x)-10
=(x²+x-2)(x²+x+5)
=(x-1)(x+2)(x²+x+5)

Answer 3

解：原式=（x²+x）²+3（x²+x）-10
=（x²+x+5）（x²+x-2）
=(x²+x+5)(x+2)(x-1)

Answer 4

(x²+x+1)(x²+x+2)-12
= (x²+x+1)^2+(x²+x+1)-12
=[(x²+x+1)+4][(x²+x+1)-3]
=(x²+x+5)(x²+x-2)
=(x²+x+5)(x+2)(x-1)