Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$ Q:Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$.My book solve it leting the roots of the equation be $\alpha,\alpha+3,\beta$ then find the equation whose roots are $\alpha-3,\alpha,\beta-3$.And i know how to find the equation whose roots are diminished by $3$ and they get it $2x^3+19x^2+53x+36=0.$
Hence $(x+1)$ is a common factor of $2x^3+x^2-7x-6=0$ and $2x^3+19x^2+53x+36=0.$And they showed all the roots are: $-1,2,\frac{-3}{2}$Now my Question is "Is there exist any easier way to solve it?" Because in this process i need a lot of work in order to find the new equation and find GCD/HCF of these two equation.Any hints or solution will be appreciated.
Thanks in advance.

Just to give another approach, if the roots are $\alpha$, $\alpha+3$, and $\beta$, then, by Vieta, we have
$$\alpha+(\alpha+3)+\beta=-1/2$$ 
and
$$\alpha(\alpha+3)+\alpha\beta+(\alpha+3)\beta=-7/2$$
The first of these can be rewritten as $2\alpha+\beta=-1/2-3=-7/2$, which means the second implies
$$\alpha^2+3\alpha+2\alpha\beta+3\beta=2\alpha+\beta$$
This rewrites as
$$\alpha^2+\alpha+(2\alpha+2)\beta=0$$
or 
$$(\alpha+1)(\alpha+2\beta)=0$$
Now either $\alpha=-1$, or else $\alpha=-2\beta$.  But if $\alpha=-2\beta$, then $2\alpha+\beta=-7/2$ implies $\beta=7/2$, which, by the Rational Root Theorem, is not a possible root for $2x^3+x^2-7x-6$, since $7\not\mid6$. So we must have $\alpha=-1$, so that $\beta=-7/2-2\alpha=-7/2+2=-3/2$, and thus the roots are $-1$, $2$, and $-3/2$.
