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 
 A: 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? :)
A: There's been some discussion about the roots and/or coefficients not being quite right, so I'll colour the suspect-but-irrelevant-to-my-point parts in red. (Well, I do have to assume one thing is preserved after as yet unidentified corrections, namely that a substitution I'll mark in blue leads to all coefficients being integers after we've rescaled the constant term.)
If we first rescale so all coefficients are integers, we find powers of $2$ dividing a lot of coefficients. Defining $\color{blue}{y:=2x}$, we want to factorize $y^4\color{red}{-4y^3-12y^2+32y}+15$, and the rational root theorem tells us the only possible rational roots are $\pm 1,\,\pm 3,\,\pm 5,\,\pm 15$. In fact $3,\,5$ work (as requested), so we can take out a factor of $y^2-8y+15$. The ratio is a quadratic $y^2+ky+1$, and the $y^3$ coefficient tells us the value of $k$, and hence the roots $x=\frac{-k\pm\sqrt{k^2-4}}{4}$. As it stands, we get $\color{red}{k=4,\,x=-1\pm\frac{\sqrt{3}}{2}}$.
