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== | ||
+ | {{iTest box|year=2007|num-b=TB1|num-a=TB3}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 20:05, 2 January 2020
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).
See also
2007 iTest (Problems) | ||
Preceded by: Problem TB1 |
Followed by: Problem TB3 | |
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