Question

x4次方 ＋ax立方 +bx－16有两个因式（x－1）（x－2）.求a,b的值.

Answer 1

（x－1）（x－2）
=x^2-3x+2
用竖式除法
- x^2 (a-3)x +(3a-11)
- __________________________
x^2-3x+2| x^4 +ax^3 +0x^2 +bx -16
x^4 -3x^3 +2x^2
- _________________________________
- | 0 (a-3)x^3 -2x^2 +bx
- (a-3)x^3 -3(a-3)x^2+2(a-3)x
- ------------------------------------------------------
- 0 (3a-11)x^2+[b-2(a-3)]x-16
- (3a-11)x^2 -3(3a-11)x +2(3a-11)
- ----------------------------------------------------------------
0
则有-16-2(3a-11)=0
2(3a-11)=-16
3a-11=-8
3a=3
a=1
[b-2(a-3)]+3(3a-11)=0
b-2(1-3)+3(3*1-11)=0
b+4-24=0
b=20