Consider the set F of all polynomials whose coefficients are in the set of 0,1 . Let q(x)=x^3+x+1 . The number of polynomials p(x) in F of degree 14 such that the product p(x) q(x) is also in F is:

Step-by-step solution

aligned & p(x) q(x)= (x¹⁴+ ) (x^3+x+1 ) & p(x)=x¹⁴ 1 case & p(x)=x¹⁴+x^2 & =10,9,8, , 10 11 case & p(x)=x¹⁴+x^ +x^ aligned . array cc =10, & =6,5,4,3,2,1,0 =9, & =5,4,3,2,1,0 =8, & =4,3,2,1,0 =7, & =3,2,1,0 =6, & =2,1,0 =5, & =1,0 =4, & =0 array 25 cases p(x)=x¹⁴+x^ +x^ +x^r . array llr =10, & =6, & r=2,1,0 =10, & =5, & r=1,0 =10, & =4, & r=0 array 6 cases . array ll =9, & =5, r=1,0 =4, & r=0 array 3 cases =8, =4, r=0 1 cases aligned & Hence, total case =1+11+28+6+3+1 =50 cases. aligned