Problem:
Let and be the unique polynomials such that
and the degree of is less than 2 . What is ?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
Let be a root of so that . It follows that
from which , but .
Note that
Since for each root of , the remainder when is divided by is .
Note that the equation above is in the form of polynomial division, with being the dividend, being the divisor, and and being the quotient and remainder respectively. Since the degree of the dividend is and the degree of the divisor is , that means the degree of the quotient is .
Note that can't influence the degree of the right hand side of this equation since its degree is either or .
Since the coefficients of the leading term in the dividend and the divisor are both , that means the coefficient of the leading term of the quotient is also . Thus, the leading term of the quotient is .
Multiplying by the divisor gives . We have our term but we have these unnecessary terms like . We can get rid of these terms by adding to the quotient to cancel out these terms, but this then gives us . Our first instinct will probably be to add , but we can't do this as although this will eliminate the term, it will produce a term. Since no other term of the form where is an integer less than will produce a term when multiplied by the divisor, we can't add to the quotient. Instead, we can add to the coefficient to get rid of the term. Continuing this pattern, we get the quotient as
The last term when multiplied with the divisor gives . This will get rid of the term but will produce the expression , giving us the dividend as . Note that the dividend we want is of the form . Therefore, our remainder will have to be in order to get rid of the term in the expression and give us , which is what we want. Therefore, the remainder is .
The problems on this page are the property of the MAA's American Mathematics Competitions