解方程:x^4+2x^3+5x^2+4x-12=0

x^4+2x^3+5x^2+4x-12
=x^4-x³+3x³-3x²+8x²-8x+12x-12
=x³(x-1)+3x²(x-1)+8x(x-1)+12(x-1)
=(x-1)(x³+3x²+8x+12)
=(x-1)(x³+2x²+x²+2x+6x+12)
=(x-1)[x²(x+2)+x(x+2)+6(x+2)]
=(x-1)(x+2)(x²+x+6)
=0
所以该方程的根为x1=1 x2=-2 x3=(-1+√23i)/2 x4=(-1-√23i)/2
其中后两个都是复根,i为虚数单位,如果题目要求只保留实根的话,后面的两个要舍去.