Problem:
Suppose a and b are real numbers. When the polynomial x3+x2+ax+b is divided by xβ1, the remainder is 4. When the polynomial is divided by xβ2, the remainder is 6. What is bβa?
Answer Choices:
A. 14
B. 15
C. 16
D. 17
E. 18
Solution:
When the polynomial x3+x2+ax+b is divided by xβ1, the remainder is 4.
In other words,
x3+x2+ax+b=(xβ1)q(x)+4
When x=1, we obtain
a+b=2
Similarly, When the polynomial is divided by xβ2, the remainder is 6.
In other words,
x3+x2+ax+b=(xβ2)r(x)+6.
When x=2, we obtain
2a+b=β6
However, we know from above that a+b=2, which gives us a=β8 and b=10.
Thus,
bβa=10β(β8)=(E) 18β.
The problems on this page are the property of the MAA's American Mathematics Competitions