2007 iTest Problems/Problem TB2
Problem
Factor completely over integer coefficients the polynomial . Demonstrate that your factorization is complete.
Solution
Note that . If
and
, then
and
. Therefore if
and
, then
. Hence
. Dividing through gives us
![$p(x)=(x^2+x+1)(x^6-x^5+2x^3-x^2+1)$](http://latex.artofproblemsolving.com/b/c/a/bca9abdcec507b7fddd9ec50cd3abf6621bbe4c8.png)
Using the Rational Root Theorem on the second polynomial gives us that are possible roots. Only
is a possible root. Dividing through gives us
![$p(x)=(x+1)(x^2+x+1)(x^5-2x^4+2x^3-x+1)$](http://latex.artofproblemsolving.com/0/0/5/005304c9ac9f680c3b3c8435b326d311cd641a2b.png)
Note that can be factored into the product of a cubic and a quadratic. Let the product be
![$x^5-2x^4+2x^3-x+1=(x^2+ax+b)(x^3+cx^2+dx+e)$](http://latex.artofproblemsolving.com/0/4/8/048769cdb4559eb34824a05468d6fbab1d2805c9.png)
We would want the coefficients to be integers, hence we shall only look for integer solutions. The following equations must then be satisfied:
Since and
are integers,
is either
or
. Testing the first one gives
We must have that
. Therefore
, or
. Solving for
and
gives
. We don't need to test the other one.
Hence we have
![$p(x)=(x+1)(x^2+x+1)(x^2-x+1)(x^3-x^2+1)$](http://latex.artofproblemsolving.com/b/7/7/b775c04682ef984d925dc218ed2e9c63ec3939cc.png)
For any of the factors of degree more than 1 to be factorable in the integers, they must have rational roots, since their degrees are less than 4. None of them have rational roots. Hence is completely factored.
Alternate Solution
We write
The factorization of is trivial once we look at the exponents modulo
; since any root
of
satisfies
, it follows that
and the cubic factor comes as a result of polynomial division.
To prove that this is a complete factorization, it suffices to note that the factors of degree greater than have no rational roots (follows from the rational root theorem and some small cases).