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March 21st, 2018, 05:57 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, 07:57 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 647 Thanks: 412 Math Focus: Dynamical systems, analytic function theory, numerics 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. Thanks from Alexis87 Last edited by skipjack; March 24th, 2018 at 03:24 AM.
 March 24th, 2018, 02:52 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 '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". Thanks from v8archie
April 14th, 2018, 09:30 AM   #4
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Quote:
 Originally Posted by SDK 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.

 January 7th, 2019, 11:08 AM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Happened across this post and was reminded of a great video: (Don't be put off by scribbling in icon. It opens from the beginning which is very clear.) Last edited by zylo; January 7th, 2019 at 11:10 AM.

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