# Difference between revisions of "2007 iTest Problems/Problem TB2"

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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 <math>p(x)</math> is completely factored. | 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 <math>p(x)</math> is completely factored. | ||

+ | |||

+ | ===Alternate Solution=== | ||

+ | We write <cmath>p(x)=x^8+x^5+x^4+x^3+x+1=</cmath> <cmath>x^5(x^3+1)+x^3(x+1)+(x+1)=</cmath> <cmath>(x+1)(x^5(x^2-x+1) + (x^3+1))=</cmath> <cmath>(x+1)(x^5(x^2-x+1)+(x+1)(x^2-x+1))=</cmath> <cmath>(x+1)(x^2-x+1)(x^5+x+1)=</cmath> <cmath>(x+1)(x^2-x+1)(x^2+x+1)(x^3-x^2+1)</cmath> | ||

+ | |||

+ | The factorization of <math>x^5+x+1</math> is trivial once we look at the exponents modulo <math>3</math>; since any root <math>\omega</math> of <math>x^2+x+1</math> satisfies <math>\omega^3=1</math>, it follows that <math>x^2+x+1 | x^5+x+1</math> 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 <math>1</math> have no rational roots (follows from the rational root theorem and some small cases). | ||

==See also== | ==See also== | ||

[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |

## Revision as of 17:28, 13 June 2013

## 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

Using the Rational Root Theorem on the second polynomial gives us that are possible roots. Only is a possible root. Dividing through gives us

Note that can be factored into the product of a cubic and a quadratic. Let the product be

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

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).