Determine number of real roots on an incomplete polynomial Let's say that I have an incomplete quartic equation with real coefficients,  which is $$x^4 - 3x^3 + ... - 10 = 0$$
And also given 2 complex roots, $a + 2i$ and $1 + bi$ where $a$ and $b$ are real numbers.
The problem asks the sum of the real roots, but firstly I don't know how to determine if the equation even has a real root. 
I can't do Rule of Signs because obviously the polynomial is incomplete. Although I can do a (heavy) assumption that $1 + bi$ is the conjugate of $a + 2i$ or not, that still would give me $0$ or $2$ real roots.
How can I know if it has a real root or not?
EDIT : I got the second term wrong, it should be $-3x^3$ and not $-3x^2$ ! I have edited the original equation. Sorry!

Edit : The OP changed $-3x^2$ to $-3x^3$.
Let $\alpha,\beta,\gamma,\omega$ be the four roots.
Then, by Vieta's formulas,
$$\alpha+\beta+\gamma+\omega=3$$
$$\alpha\beta\gamma\omega=-10$$
Case 1 : If $(a,b)=(1,-2)$, then we may suppose that $\alpha=1+2i,\beta=1-2i$, so
$$\gamma+\omega=1,\quad \gamma\omega=-2$$
So, $\gamma,\omega$ are the solutions of $x^2-x-2=(x-2)(x+1)=0$, i.e.
$$\gamma,\omega=2,-1$$
giving that sum of real roots is $\color{red}{1}$.
Case 2 : If $(a,b)\not=(1,-2)$, then there is no real roots since four roots are $a\pm 2i,1\pm bi$ where $b\not=0$. However, as Tobias Kildetoft comments, this does not happen because there is no real a,b such that $(a^2+2^2)(1^2+b^2)=-10$. 
