4x³-12x²+20x-12=0x³-3x²+5x-3=0x³-3x²+3x-1+2x-2=0(x-1)³+2(x-1)=0(x-1)[(x-1)²+2]=0(x-1)(x²-2x+1+2)=0(x-1)(x²-2x+3)=0楼主关注待定系数法,那……就用待定系...系数不等于1怎么写还是3x³-2x+1=0，像这个怎么用待定系数法，求解设：3x³-2x+1=(3x+a)(x²+bx+c)或者设：3x³-2x+1=(x+a)(3x²+bx+c)