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.

Here is one more approach. First, we do not compute anything, just make sentences. One of the paired roots is $a$, say, the other one is $a\pm 3$, and there is one more root, such that the sum of all three is (Vieta) a rational number, well $-1/2$, but it is not so important, it is rational, this is important. So all three roots can be expressed $\Bbb Q$-linearly (or rather affine) using the one root $a$. The "quadratic" Vieta relation delivers thus a quadratic equation for $a$ with rational coefficients. We use division with rest (algorithm of Euclid, roots also annihilate the rest if not zero) of the given polynomial
$$
P = 2x^3+x^2-7x-6
$$
with this "quadratic equation". If we have some (non zero) rest, it has degree one, so $a$ is rational. If there is no rest, the quotient gives an other rational root. 
In both cases, we know now that $P$, polynomial with integer coefficients, has a rational root. The rational roots are (in similar cases) always of the shape
$$
\pm\frac
{\text{divisor of the free coefficient}}
{\text{divisor of the leading coefficient}}\ ,
$$
so in our cases we have to try the values
$$
\pm\frac
{\text{$1$ or $2$ or $3$ or $6$}}
{\text{$1$ or $2$}}\ .
$$
Many values? Not so many if we also use some further information, the continuity of the function. We "calculate" $P(0)=-6$, and thus there is at least one root between $0$ and $\infty$. (It may be rational or not so far. We speculate there is at most $6$ a rational root, so we try to take "the half", we compute $P(2)$ and/or $P(3)$.) Well, i compute $P(3)$ unluckily, get $P(3)= 36>0$, so there is one root between $0$ and $3$. We try all values from the list that match, $1$, $3/2$ and $2$, using best the Horner scheme. OK, $2$ is a root. We use the Horner scheme (below) to also make polynomial division, get $2x^2 + 5*x + 3$, and can use the formula, or also the same (now speculative) search for rational roots. Yes, $-1$ is a root, and the remained one is $-3/2$ (Vieta, product is $3/2$).
The Horner scheme is simple:
$$
\begin{array}{r|rrrr}
  & 2 & 1 & -7 & -6\\\hline
2 & 2 &2\cdot 2+1=5 & 2\cdot 5-7=3 & 2\cdot 3-6=\boxed0\\
-1& 2 & (-1)\cdot 2+5 = 3&(-1)\cdot 3+3  =\boxed0 &\\
-3/2 & 2 & \boxed0 &&
\end{array}
$$
Well, usually one sees only
$$
\begin{array}{r|rrrr}
  & 2 & 1 & -7 & -6\\\hline
2 & 2 & 5 & 3 & \boxed0\\
-1& 2 & 3&\boxed0 &\\
-3/2 & 2 & \boxed0 &&
\end{array}
$$
instead.
The roots are on the left hand side, the rests w.r.t. the corresponding divisions are boxed.
