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.

$$2x^3+x^2-7x-6$$ and $$2x^3+19x^2+53x+36$$
have a common root, which is such that
$$x^2-7x-6=19x^2+53x+36.$$
By solving the quadratic, this root is one of $-1$ or $-\dfrac73$. Then $-1$ fits and by long division you reduce to 
$$2x^2-x-6=0.$$
