Polynomial Rings and Irreducibility Criteria MCQ Quiz in मल्याळम - Objective Question with Answer for Polynomial Rings and Irreducibility Criteria - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Solution -  

Option 1) 

Given, polynomial \(x^2+x+1\) has no root in \(Z_2\) 

so it is irreducible .

Option 2) 

Given, polynomial \(x^2 - 2\) has no root in Q 

so it is irreducible 

Option 3) 

As every polynomial of odd degree 

has atleast one real root in R so it is reducible.

Option 4) 

\(1+x+\frac{x^2}{2 !}+\cdots+ \frac{x^{101}}{101!}\)

\(\frac{x^{101}+101x^{100}+....+101!}{101!}\)

let q(x) = \(x^{101}+101x^{100}+...+ 100! \) 

taking p= 101 here p divides \(a_o,a_1, a_{n-1} \ and \ p \ does \ not \ divide \ a_n, a_o\) 

then , By einstein Criteria Q[x] is irreducible polynomial. 

Therefore, Correct Option (s) are Option 1), 2) and 4).