Solve in $\mathbb R$ : $x^4-2x^{3}-3x^{2}+4x+\frac{15}{16}=0$ Problem : 
Solve in $\mathbb R$ : 
$x^4-2x^{3}-3x^{2}+4x+\frac{15}{16}=0$
The roots are : (by Wolfram Alpha ) 
$x_{i}=\frac{5}{2},\frac{3}{2},-\frac{1+\sqrt 2}{2},-\frac{1-\sqrt 2}{2}$ 
I don't know how to start because in the first place it is not simple to try roots like $\frac{3}{2}$ 
So I need some ideas or hints to factorization this cubic equation 

Suppose that instead of a polynomial in $x$ we had a polynomial in $2x$.
$$\begin{align}
x^4 - 2x^3-3x^2+4x+\frac{15}{16}&=0\\
16x^4 - 16\cdot2x^3-16\cdot3x^2+16\cdot4x+15&=16\cdot0\\
(2x)^4 - 4(2x)^3-12(2x)^2+32(2x)+15&=0\\
y^4 - 4y^3-12y^2+32y+15&=0\\
\end{align}$$
Where $y=2x$.
Can you take it from here? :)
