Find the coefficients of this polynomial. Given $f(x)=x^4 +ax^3 +bx^2 -3x +5$ , $  a,b\in \mathbb{R}$ and knowing that $i$ is a root of this polynomial. what are $a,b$? 
My work until now: since all the coefficients are real, I now know that $i$ and $-i$ are roots of this polynomial, there must be two more roots. I tried to use vietta but it didn't really help, I get that $x_3+x_4=-a$ and $i(-i)x_3x_4=5$. What am I missing? 
Appreciate any help.

By long division,
$$
\require{enclose}
\begin{array}{r}
x^2+ax+(b-1)\\[-4pt]
x^2+1\enclose{longdiv}{x^4+ax^3+bx^2-3x+5}\\[-4pt]
\underline{x^4\phantom{\ +ax^3}+x^2\ }\phantom{-3x+5\ \ }\\[-2pt]
ax^3+(b-1)x^2-3x\phantom{\,+5\ \ }\\[-4pt]
\underline{ax^3\phantom{\,\,+(b-1)x^2}+ax}\phantom{\ +5}\\[-2pt]
(b-1)x^2-(a+3)x\ +5\\[-4pt]
\underline{(b-1)x^2\phantom{\ \qquad}+(b-1)}\\[-2pt]
-(a+3)x+(6-b)
\end{array}
$$
We must have that $-(a+3)x+(6-b)=0$. Therefore, $b=6\text{ and }a=-3$. Thus,
$$
\left(x^2+1\right)\left(x^2-3x+5\right)=x^4-3x^3+6x^2-3x+5
$$
