How to factor $x^4 +3x -2$? I have figured out there is two roots between $0$ and $1 ,-1$ and $-2$ for $x^4 +3x -2 = 0$.
Therefore there should be two factors $(x + a)$ and $(y - b)$ where $a,b \in R^+$. 
But how to find these $a$ and $b$?
When they found I can find the next factor in $ax^2+bx+c$ form and can check for further factors easily.

For it to have some "nice" linear factors, the roots must be one of $\pm 1,\pm 2$ (this is due to the rational root theorem). You can quickly check that these are not the roots. The next bet is quadratic factors, i.e.,
$$(x^4+3x-2) = (x^2+ax+b)(x^2+cx+d)$$
Expanding the right hand side gives us
\begin{align}
a+c & = 0\\
b + d + ac & = 0\\
ad+bc & = 3\\
bd & = - 2
\end{align}
This gives us the factors to be
$$(x^2-x+2) \text{ and } (x^2+x-1)$$
