\(x^4+x^3+3x^2+ax+4=\left(x^2-x+b\right)\left(x^2+cx+\frac{4}{b}\right)\)
\(=x^4+\left(c-1\right)x^3+\left(\frac{4}{b}+b-c\right)x^2+\left(bc-\frac{4}{b}\right)x+4\)
=> c -1 = 1 => c =2
=> 4/b +b -2 =3 => b2 -5b +4 =0 => b =1 hoặc b =4
+ Nếu b =1 => a = bc - 4/b =1.2 - 4/1 = -2
+ Nếu b =4 => a =............ = 4.2 - 4/4 = 7
Vậy a = -2 khi b =1
a = 7 khi b =4