When the polynomial x2007 + 1 is divided by x2 – 3x +2, the remainder can be expressed in the form ax+b. Find the value of a.
My Solution:
Let k be the quotient, then:
x2007 + 1 = k(x2 – 3x +2) + (ax+b)
Since x2 – 3x +2 = (x-2)(x-1),
x2007 + 1 = k(x-2)(x-1) + (ax+b)
To omit k and to solve for the value of a easily, set x=2 and x=1:
For x=2
(2)2007 + 1= k(2-2)(2-1) + (2a+b)
22007 + 1= 2a+b (*)
For x=1
(1)2007 + 1 = k(1-2)(1-1) + (a+b)
2 = a+b (**)
To solve for the value of a, perform elimination using (*) and (**).
..Which in turns give: a=22007-1
$interesting$
Great! But, since you didn’t explicitly mention the Remainder Theorem, I thought there would be no harm in mentioning it. Your solution actually contains a “proof” of the Remainder Theorem for a special case that can be easily generalized.