116The middle equation givesa(c−b)=0, so that eithera=0 orb=c. In thefirst case, we obtainc+b=10cb=1.Substitutingc=b−1, thefirst equation givesb2−10b+1=0.But the quadratic formula givesb=5±2√6, which is irrational.On the other hand,ifb=c, thenbc=1 impliesb= ±1=c. Thefirst equation givesa2= −10±2<0, and this is also impossible.We conclude that there is no factorization off(x)inQ[x].(ix)f(x)=x6−210x−616.Solution.Eisenstein’s criterion applies, for 7|210 and 7|616,but 72616.(x)f(x)=350x3+x2+4x+1.Solution.Reducing mod 3 to gives an irreducible cubic inF3[x].3.88Ifpis a prime, prove that there are exactly13(p3−p)monic irreduciblecubic polynomials inFp[x].Solution.Absent.3.89Prove that there are exactly 6 irreducible quintics inF2[x].Solution.There are 32 quintics inF2[x], 16 of which have constant term 0;that is, have 0 as a root. Of the 16 remaining polynomials, we may discard
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