It's clear that $a^2 - b^2$ has, and that $a^2 + b^2$ doesn't have. But when a polynomial gets longer and longer, do you have any sort of rule, like you have with natural numbers (when they end in $2$ are even, in $5$ are divisible by $5$, and so on).
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In general, the answer to this question will depend on the base ring of the coefficients. Have you come across the concept of a ring before? Are you only interested in the case where all of the factors have integer coefficients? For example, every polynomial factors fully into linear factors with complex coefficients. – Joshua Pepper Mar 08 '14 at 16:49
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Also, you need to be slightly more clear about what types of polynomials you're considering. Does your polynomial have two variables, $a$, and $b$, or should we think of this as a polynomial of the form $X^2-b^2$, where $b$ is fixed, and $X$ is an indeterminate (a variable)? – Joshua Pepper Mar 08 '14 at 16:54
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Start with the Eisenstein criterion for irreducible polynomials, then study the Zassenhaus algorithm for factorization of univatiate polynomials with integer coefficients. Also, the LLL algorithm can be used to meaningfully guess polynomial factors. – Lutz Lehmann Mar 09 '14 at 17:21
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We can compare polynomials to numbers and study irreducible polynomials in a manner analogous to how we study prime numbers, and this is often done in abstract algebra.
Assuming that you are more precisely asking about how to tell if a quadratic polynomial is reducible over the real numbers $\mathbb{R}$, you simply check the discriminant. Any quadratic $ax^2+bx+c$, is reducible over the reals if and only if $b^2-4ac>0$.
So if $a$ is variable and $b$ is constant, $0^2-4(1)(-1)>0 \implies a^2-b^2$ is reducible over the $\mathbb{R}$. Similarly, $0^2-4(1)(1)<0 \implies a^2+b^2$ is irreducible over $\mathbb{R}$.
As an aside, you can also check if an integer-valued polynomial is reducible over $\mathbb{Q}$ using Gauss's Lemma: "When a polynomial is integer valued, one may appeal to Gauss's lemma which states that if the coefficients of a non-constant polynomial $f$ are relatively prime and $f$ is irreducible in $\mathbb{Z}[X],$ then $f$ is irreducible in $\mathbb{Q}[X].$" (from an old post)
William Chang
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Every polynomial has factors $a^2+b^2=(a+ib)(a-ib)$ Note $i=\sqrt{-1}$ anyway there aren't much info on whether a polynomial has real roots, Sturm's theorem,anyway odd degree polynomials always have real roots.For rational roots
