Can a Non-Factorable Function Have Rational Real Roots?

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A non-factorable function can have rational real roots, as demonstrated by the example f(x) = x + a, where a is a rational number. However, for polynomials of degree greater than one that cannot be expressed as a product of lower-degree polynomials with rational coefficients, rational roots are not possible. If a polynomial has a rational root, it can be factored into a linear component, contradicting its non-factorable nature. Therefore, while some non-factorable functions can have rational roots, most polynomials of higher degrees cannot. This highlights the relationship between rational roots and polynomial factorization.
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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 f(x):= x+a where a \in \mathbb{Q}.
 
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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