已知多項(xiàng)式2x的四次方-3x的三次方+ax的平方+7x+b能被(x-1)(x+2)整除
2x^4-3x^3+ax^2+7x+b=(x-1)(x+2)(2x^2+cx+d)
2x^4-3x^3+ax^2+7x+b=(x^2+x-2)(2x^2+cx+d)
2x^4-3x^3+ax^2+7x+b=2x^4+cx^3+dx^2+2x^3+cx^2+dx-4x^2-2cx-2d
2x^4-3x^3+ax^2+7x+b=2x^4+(c+2)x^3+(d+c-4)x^2+(d-2c)x-2d
(c+2)=-3(d+c-4)=a(d-2c)=7-2d=b
解得
c=-5d=-3a=-12 b=6