relation between roots and coefficient in a cubic polynomial

Question

If $\alpha,\beta,\gamma$ are roots of the cubic equation $$2x^{3}+3x^2-x-1=0$$ then I want to find the equation whose roots are $\frac{\alpha}{\beta+\gamma}, \frac{\beta}{\gamma+\alpha}, \frac{\gamma}{\alpha+\beta}$.

I have $\alpha+\beta+\gamma$= - $\frac{3}{2}$.

Therefore roots become $\frac{\alpha}{-\frac{3}{2} -\alpha}$,$\frac{\beta}{-\frac{3}{2} -\beta}$, $\frac{\gamma}{-\frac{3}{2} -\gamma}$.

To find the equation we have to find their sums and products, but it looks like complicated.

Is their any easy trick ?

Answer 1

If $$ 2x^3+3x^2-x-1=0 $$ has roots $\alpha,\beta,\gamma$, then substituting $x\mapsto\frac1x$ (and multiplying by $-x^3$ to clear denominators) $$ \begin{align} &-x^3\left(\frac2{x^3}+\frac3{x^2}-\frac1x-1\right)\\ &=x^3+x^2-3x-2=0 \end{align} $$ has roots $\frac1\alpha,\frac1\beta,\frac1\gamma$. Then substituting $x\mapsto-\frac23x$ (and multiplying by $-\frac{27}8$ to clear denominators) $$ \begin{align} &-\frac{27}8\left(\left(-\frac23x\right)^3+\left(-\frac23x\right)^2-3\left(-\frac23x\right)-2\right)\\ &x^3-\frac32x^2-\frac{27}4x+\frac{27}4=0 \end{align} $$ has roots $\frac{-3/2}\alpha=\frac{\alpha+\beta+\gamma}\alpha,\frac{-3/2}\beta=\frac{\alpha+\beta+\gamma}\beta,\frac{-3/2}\gamma=\frac{\alpha+\beta+\gamma}\gamma$. Next, substituting $x\mapsto x+1$ $$ \begin{align} &(x+1)^3-\frac32(x+1)^2-\frac{27}4(x+1)+\frac{27}4\\ &=x^3+\frac32x^2-\frac{27}4x-\frac12=0 \end{align} $$ has roots $\frac{\beta+\gamma}\alpha,\frac{\alpha+\gamma}\beta,\frac{\alpha+\beta}\gamma$. Finally, substituting $x\mapsto\frac1x$ (and multiplying by $-4x^3$ to clear denominators) $$ \begin{align} &-4x^3\left(\frac1{x^3}+\frac32\frac1{x^2}-\frac{27}4\frac1x-\frac12\right)\\ &=2x^3+27x^2-6x-4=0 \end{align} $$ has roots $\frac\alpha{\beta+\gamma},\frac{\beta}{\alpha+\gamma},\frac{\gamma}{\alpha+\beta}$.

Answer 2

First, note that $\beta+\gamma = -\frac{3}{2} -\alpha$.

Now, let $y=\frac{x}{-\frac{3}{2} -x}=g(x)$. Then $x=h(y)= -\frac{3 y}{2 (y + 1)}$.

Therefore, if $z$ is a root of $f(x)$, then $w=g(z)$ is a root of $j(w)=f(h(w))$. Finally, clear the denominators in $j(w)=0$ to get a polynomial equation for $w$.

Answer 3

While this is not a general method for zeroes of a cubic polynomial, this particular polynomial is amenable to familiar methods. We can apply the Rational Zeroes Theorem to find that one zero is $ \ \alpha \ = \ -\frac12 \ \ , $ so we may factor $ \ 2x^3 + 3x^2 - x - 1 \ = \ \left(x + \frac12 \right)·2·(x^2 + x - 1) \ = \ 0 \ \ . $ This polynomial also has the helpful property that it has symmetry about $ \ x \ = \ -\frac12 \ \ : \ 2·\left(u - \frac12 \right)^3 + 3·\left(u - \frac12 \right)^2 - \left(u - \frac12 \right) - 1 $ $ = \ 2u·\left(u^2 - \frac54 \right) \ \ . $ The other zeroes of the original polynomial are then $ \ \beta \ , \ \gamma \ = \ \alpha \ \pm \ \Delta \ = \ -\frac12 \pm \frac{\sqrt5}{2} \ \ . \ $ (We will have need of $ \ \Delta^2 \ = \ \frac54 \ $ shortly. These two zeroes are in fact $ \ \frac{1}{\phi} \ $ and $ \ -\phi \ \ , $ with $ \ \phi \ $ being the "Golden Ratio", but we won't be making use of that in this discussion.)

We seek a cubic polynomial with the transformed zeroes $$ \alpha' \ = \ \frac{\alpha}{\beta \ + \ \gamma} \ = \ \frac{-1/2}{-1} \ = \ \frac12 \ \ \ , \ \ \ \beta' \ = \ \frac{\beta}{\alpha \ + \ \gamma} \ = \ \frac{\alpha \ + \ \Delta}{2\alpha \ - \ \Delta} \ = \ \frac{1 \ + \ 2\Delta}{2 \ - \ 2\Delta} \ \ , $$ $$\gamma' \ = \ \frac{\gamma}{\alpha \ + \ \beta} \ = \ \frac{\alpha \ - \ \Delta}{2\alpha \ + \ \Delta} \ = \ \frac{1 \ - \ 2\Delta}{2 \ + \ 2\Delta} \ \ . $$

From these, we compute the coefficients of a new monic polynomial from $$ c' \ \ = \ \ -(\alpha' · \beta' · \gamma') \ \ = \ \ -\frac{(1 \ + \ 2\Delta)·(1 \ - \ 2\Delta)}{2 ·(2 \ - \ 2\Delta)·(2 \ + \ 2\Delta)} \ \ = \ \ \frac{4\Delta^2 \ - \ 1}{8 · (1 \ - \ \Delta^2)} $$ $$ = \ \ \frac{5 \ - \ 1}{-2} \ \ = \ \ -2 \ \ ; $$ $$ a' \ \ = \ \ -(\alpha' \ + \ \beta' \ + \ \gamma') \ \ = \ \ - \left[ \ \frac12 \ + \ \frac{1 \ + \ 2\Delta}{2 \ - \ 2\Delta} \ + \ \frac{1 \ - \ 2\Delta}{2 \ + \ 2\Delta} \ \right] \ \ = \ \ \frac32· \left(\frac{\Delta^2 \ + \ 1}{ \Delta^2 \ - \ 1} \right) $$ $$ = \ \ \frac32· \left(\frac{5 \ + \ 4}{ 5 \ - \ 4} \right) \ \ = \ \ \frac{27}{2} \ \ ; $$ $$ \frac{b'}{c'} \ \ = \ \ \frac{\alpha' · \beta' \ + \ \alpha' · \gamma' \ + \ \beta' · \gamma'}{- \ (\alpha' · \beta' · \gamma')} \ \ = \ \ - \left(\frac{1}{\alpha'} \ + \frac{1}{\beta'} \ + \ \frac{1}{\gamma'} \right) $$ $$ = \ \ - \left[ \ 2 \ + \ \frac{2 \ - \ 2\Delta}{1 \ + \ 2\Delta} \ + \ \frac{2 \ + \ 2\Delta}{1 \ - \ 2\Delta} \ \right] \ \ = \ \ \frac{6}{ 4\Delta^2 \ - \ 1} \ \ = \ \ \frac{6}{ 5 \ - \ 1} \ \ = \ \ \frac32 $$ $$ \Rightarrow \ \ b' \ \ = \ \ \frac32 \ · \ (-2) \ \ = \ \ -3 \ \ . $$ A cubic polynomial having the transformed zeroes which has integer coefficients is then $ \ 2x^3 \ + \ 27x^2 \ - \ 6x \ - \ 4 \ \ . $