Can a Non-Factorable Function Have Rational Real Roots?

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SUMMARY

A non-factorable function can indeed have rational real roots, as demonstrated by the function f(x) = x + a, where a is a rational number (a ∈ ℚ). However, for polynomials of degree greater than one with rational coefficients that cannot be expressed as the product of lower-degree polynomials, rational roots do not exist. This is due to the fact that any rational root a of such a polynomial implies the existence of a linear factor (x - a), leading to a contradiction of the non-factorability condition.

PREREQUISITES
  • Understanding of polynomial functions and their properties
  • Knowledge of rational numbers and their representation
  • Familiarity with the concept of factorization in algebra
  • Basic grasp of polynomial degree and roots
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  • Study the Rational Root Theorem and its implications for polynomial equations
  • Explore the concept of irreducible polynomials over the rationals
  • Investigate examples of non-factorable polynomials and their characteristics
  • Learn about polynomial factorization techniques and their limitations
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Miracles
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can a non-factorable function has rational real roots?
 
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Miracles said:
can a non-factorable function has rational real roots?
Yes, for example take [itex]f(x):= x+a[/itex] where [itex]a \in \mathbb{Q}[/itex].
 
Are you asking if a polynomial with rational coefficients which cannot be written as the product of polynomials with rational coefficients of smaller degree can have rational roots? Except for the trivial case of polynomials of degree one, the answer is no, because any rational root a of f(x) translates to a linear factor (x-a) of f(x), ie, there is a polynomial g(x) with rational coefficients such that f(x)=(x-a)g(x).
 

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