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.
 A: $$2x^3+x^2-7x-6$$ 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 
$$2x^2-x-6=0.$$
A: 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 $-3/2$.
A: 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.
