已知多项式2x的四次方-3x的三次方+ax的平方+7x+b能被(x-1)(x+2)整除,求a,b的值

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=-5 d=-3 a=-12 b=6