Question

(x^2+3x-1)(x^2+3x-2)-6 因式分解

Answer 1

(x^2+3x-1)(x^2+3x-2)-6=0，先猜根，因为明显是3×2-6=0，所以观察得到x=1时等式成立所以有一个因式是（x-1），所以【（x-1）^2+5x-2】【（x-1）^2+5x-3】-6=（x-1）^4+(10x-5)(x-1)^2+25x^2-25x+6-6=(x-1)[(x-1)^3+(10x-5)(x-1)+25x].............后面自己岔开化吧！

Answer 2

(x²+3x-1)(x²+3x-2)-6
=(x²+3x)²-3(x²+3x)+2-6
=(x²+3x)²-3(x²+3x)-4
=(x²+3x-4)(x²+3x+1)
=(x+4)(x-1)(x²+3x+1)

Answer 3

(x²+3x)²-3(x²+3x)-4
=(x²+3x-4)(x²+3x+1)
=(x+4)(x-1)(x²+3x+1)
在有理数范围内到此为止,在实数范围内则继续.

Answer 4

(x^2+3x-1)(x^2+3x-2)-6
=(x^2+3x-1)[(x^2+3x-1)-1]-6
=(x^2+3x-1)^2-(x^2+3x-1)-6

=(x^2+3x-1-3)(x^2+3x-1+2)

=(x^2+3x-4)(x^2+3x+1)

=(x+4)(x-1)(x^2+3x+1)

Answer 5

[(x^2+3x)-1][(x^2+3x)-2]-6
=(x^2+3x)^2-3(x^2+3x)+2-6
=x^4+6x^3+6x^2+9x-4