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Irreducible-Polynomial-Examples2

Q1) You know that x^2+1 is irreducible polynomial in Q[x] and   therefore                Q[x]/x^2+1 is a body. Find the following :

(a) [2x+1][3x+7] in Q[x]/x^2+1

(b) Find the inverse of [4x+3] in Q[x]/x^2+1.


Solution:

          Q[x]/x^2+1 = { ax+b +(x^2+1)/ a, b \(\epsilon\) Q }

                                 ={ [ax+b]/ a, b \(\epsilon\)Q }

                     [x^2+1]=0 then [x^2]=[-1]

 (a) [2x+1][3x+7] = [6x^2+17x+7]

                                 =[6[-1]+17x+7]

                                 =[-6+17x+7]

                                 =[17x+1]


(b) Let the inverse of [4x+3] is [ax+b]

     then [4x+3][ax+b]=1

               [4ax^2+(3a+4b)x+3b]=1

               [4a(-1)+(3a+4b)x+3b]=1

               [(3a+4b)x +(-4a+3b)]=1

               3a+4b =0 and -4a + 3b =1

           solving  them we get a= -4/25 , b=3/25

Therefore the inverse of [4x+3] is [\(\frac{-4x}{25}+\frac{3}{25}\)].






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