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Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite Let $f(x) = x^12+2x^6-2x^3+2$ and let $K$ be it's splitting field. Is the Galois group of $K/Q$ solvable? Since we're really only interested in the polynomial $x^4+2x^2-2x+2$, and we know that there exists a quartic formula. It is enough to conclude that the polynomial is solvable by radicals and therefore the Galois group is solvable. Is this solution correct? The question also asks if this polynomial is irreducible. This is easy to show via Eisenstein's criterion. But is the irreducibility of the polynomial important in any way for showing it is solvable? Feb 2016 abstract-algebra proof-verification galois-theory share | cite | improve this question edited Aug 12 at 9:34 Bernard 111k 6 35 103 asked Aug 12 at 8:31 iYOA 605 4 9 1 Irreducibility is not needed since when you compute the roots in radicals you compute them all. –  user583012 Aug 12 a

Clash Royale CLAN TAG #URR8PPP up vote -1 down vote favorite Imagine a car travelling on a straight road at speed $u$ metres per second. The driver sees a kangaroo ahead and brakes to stop, with a reaction time of two seconds. In such circumstances, the distance covered $x$ seconds after seeing the kangaroo is given by: $$d(x) = begincases ux, & textif $0 leq x leq 2$,\ ux - 2(x - 2)^2, & textif $2 leq x leq 2 + fracu4$. endcases $$ Let $f: [0,infty) to [0,infty)$ be such that $f(u)$ is the total braking distance before the car comes to a complete standstill corresponding to the initial speed $u$. Find a formula for $f$. I've tried but just have no ideas what I have to do. Can anyone help me please!! functions share | cite | improve this question edited Aug 12 at 8:37 asked Aug 12 at 8:27 Mike LoongBoong 6 3 1 I think your first $f(x)$ should have a different name such as $d(x)$. Otherwise you are using $f$ as t