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.
 A: Alternative (but poor) solution:
As we know that $\pm i$ are roots, $x^2+1$ is a factor. Then by long division we obtain the remainder
$$-(a+3)x+(-b+6).$$
But this remainder must vanish, and $a=-3,b=6$.
A: $$f(x)=x^4 +ax^3 +bx^2 -3x +5$$
Since $i$ is a root, we have
$$f(i) = 0$$
$$\implies 1 -ia -b -3i + 5 = 0$$
$$\implies (6-b) + i (-a -3) = 0$$
Since $a, b \in \mathbb{R}$, the real and imaginary part of the above expression must separately vanish.
Hence $$-a -3 = 0$$ and $$6 - b = 0$$
Finally
$$a = -3, b = 6$$
Note
In general, if $a = a_x + i a_y$ and $b = b_x + i b_y$, then we may have many solutions that satisfy $a_y - b_x + 6 = 0, a_x + b_y + 3 = 0$. Since $a, b \in \mathbb{R}$, we must have $a_y = b_y = 0$
A: 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,
$$
\left(x^2+1\right)\left(x^2-3x+5\right)=x^4-3x^3+6x^2-3x+5
$$
A: $$f(i)=i^4 + a i^3 + b i^2 - 3 i + 5=-i a-b+(6-3 i)=0$$
$$6-b-i(3+a)=0$$
as $a,b\in\mathbb{R}$ we have $6-b=0,3+s=0$
$a=-3,b=6$
