已知多项式2X^4+AX^3+BX^2+10X+4有两个因式(X^2+2X+1),(X+2),求A+B

设别的因式是Y则2X^4+AX^3+BX^2+10X+4=Y(X^2+2X+1)(X+2)2X^4+AX^3+BX^2+10X+4=Y(X+1)^2(X+2)X=-1,则X+1=0所以右边=0所以左边也等于0X=-1,2X^4+AX^3+BX^2+10X+4=2-A+B-10+4=-A+B-4=0A-B=-4 (1)同理X=-2,X+2=0,所以右...