Question

已知多项式2x^4+3x^3+ax^2+7x+b能被x^2+x-2整除，求a b的值

Answer 1

∵多项式2x^4+3x^3+ax^2+7x+b能被x^2+x-2整除，且x^2+x-2=(x+2)(x-1)
∴2x^4+3x^3+ax^2+7x+b分解因式中含有因式(x+2)和(x-1)
令f(x)=2x^4+3x^3+ax^2+7x+b
则f(1)=0 且 f(-2)=0
即12+a+b=0
-6+4a+b=0 联立解得：a=6,b=-18

Answer 2

多项式2x^4+3x^3+ax^2+7x+b能被x^2+x-2整除
则有2X^4+3X^3+AX^2+7X+B=(X^2+X-2)(CX^2+DX+E)
=CX^4+CX^3-2CX^2+DX^3+DX^2-2DX+EX^2+EX-2E
必然有
C=2
C+D=3 ==>D=1
-2C+D+E=A
-2D+E=7 ==>E=9
-2E=B ==>B=-18
==>A=-4+1+9=6

Answer 3

C=2
C+D=3 ==>D=1
-2C+D+E=A
-2D+E=7 ==>E=9
-2E=B ==>B=-18
==>A=-4+1+9=6