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March 21st, 2018, 06:57 PM   #1
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Irreducible or reducible polynomial?

Hi all, I have used rational root test and obtain this result for the question:

f(x) = 2x^4+8x^3+5x^2−7x−3=(2x^2+2x−3)(x^2+3x+1)

This shows that there is no linear factor but a quadratic factors instead. Do I still consider such polynomial as reducible or irreducible?

Am I correct to say that f(x) has no linear factors, hence (2x^2+2x−3) and (x^2+3x+1) have no linear factors. As (2x^2+2x−3) and (x^2+3x+1) are of degree 2, it follows that they are irreducible in Q[x]?
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March 21st, 2018, 08:57 PM   #2
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It is correct as long as you say it is irreducible over $\mathbb{Q}[x]$ because it has no linear factors in $\mathbb{Q}[x]$. It does have linear factors; every polynomial splits into linear factors over its splitting field. When discussing irreducibility, it is important to be specific about which fields/rings you are referring to. The property of being irreducible is meaningless without specifying a field/ring.

It's also worth pointing out, since it's a common mistake, that a polynomial can be reducible over some polynomial ring even if it has no linear factors in that ring. Linear factors are only special when dealing with quadratic polynomials or polynomials of prime degree.
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Last edited by skipjack; March 24th, 2018 at 04:24 AM.
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March 24th, 2018, 03:52 AM   #3
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'Reducible' does NOT mean "can be factored into linear factors". It simply means "can be factored into a product of polynomials of lower degree than the original polynomial. You are correct that this polynomial cannot be factored (over Q) into linear terms but the fact that it can be factored into the product of two quadratic polynomials means that it is "reducible".
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April 14th, 2018, 10:30 AM   #4
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Quote:
Originally Posted by SDK View Post
It's also worth pointing out, since it's a common mistake, that a polynomial can be reducible over some polynomial ring even if it has no linear factors in that ring. Linear factors are only special when dealing with quadratic polynomials or polynomials of prime degree.
In fact, just quadratic or cubic. If your polynomial $f \in R[X]$ has degree at least four, we can write $\deg f = a + b$ for some integers $a,b \geq 2$. Then, even if $f$ doesn't have any linear factors in $R[X]$, it might still factor as $gh$ for some polynomials $g$ and $h$ in $R[X]$ of degrees $a$ and $b$, respectively.
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