Solve in $\mathbb R$ : $x^4-2x^{3}-3x^{2}+4x+\frac{15}{16}=0$

Question

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

Answer 1

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? :)

Answer 2

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 $\color{red}{k=4,\,x=-1\pm\frac{\sqrt{3}}{2}}$.