How do i solve the polynomial equation?

Question

Let the roots of the equation: $2x^3-5x^2+4x+6$ be $\alpha,\beta,\gamma$

1. State the values of $\alpha+\beta+\gamma,\alpha\gamma+\alpha\beta+\beta\gamma,\alpha\beta\gamma$
2. Hence, or otherwise, determine an equation with integer coefficients which has $\frac{1}{\alpha^2}\frac{1}{\beta^2}\frac{1}{\gamma^2}$

For Question 1 I let the roots equal: $(x-\alpha)(x-\beta)(x-\gamma)$ which equals: $x^3-x^2(\alpha+\beta+\gamma)+x(\alpha\gamma+\alpha\beta+\beta\gamma)-\alpha\beta\gamma$

I then equated it as:

$\alpha+\beta+\gamma =\frac{5}{2}$

$\alpha\gamma+\alpha\beta+\beta\gamma=2$

$\alpha\beta\gamma=-6$

Answering question 2 I went and did:

$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}$ which equal $\frac{\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2}{\alpha^2\beta^2\gamma^2}=\frac{(\alpha\beta)^2+(\alpha\gamma)^2+(\beta\gamma)^2}{(\alpha\beta\gamma)^2}$

that would give me the sum of the roots and

$\frac{1}{(\alpha\beta\gamma)^2}$ would give me the product of the roots but kinda confused as to how to finish this question.

Answer 1

All this is about Vieta's relations between the elementary symmetric funnctions of the roots of a polynomial and its coefficients. Let's denote $$s=\alpha+\beta+\gamma,\quad q=\alpha\beta+\beta\gamma+\gamma\alpha, \quad p=\alpha\beta\gamma.$$ Thus you have to find the values of

• $S=\dfrac1{\alpha^2}+\dfrac1{\beta^2}+\dfrac1{\gamma^2}=\dfrac{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2}{(\alpha\beta\gamma)^2}=\dfrac{(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)}{(\alpha\beta\gamma)^2}=\dfrac{q^2-2sp}{p^2}$.
• $Q=\dfrac1{\alpha^2\beta^2}+\dfrac1{\beta^2\gamma^2}+\dfrac1{\gamma^2\alpha^2}=\dfrac{\alpha^2+\beta^2+\gamma^2}{(\alpha\beta\gamma)^2}=\dfrac{s^2-2q}{p^2}$.
• $P=\dfrac1{\alpha^2}\dfrac1{\beta^2}\dfrac1{\gamma^2}=\dfrac1{p^2}$.

Answer 2

I think you're asking how to find the numerator of the fraction giving the other values.

Notice that:

$$(\alpha \beta + \alpha \gamma + \beta \gamma)^2 = (\alpha \beta)^2 + (\alpha \gamma)^2+(\beta \gamma)^2 + (\alpha + \beta + \gamma)(2)(\alpha \beta \gamma)$$ which you can find by simply expanding and factoring

Therefore,

$$(\alpha \beta)^2 + (\alpha \gamma)^2+(\beta \gamma)^2 = (\alpha \beta + \alpha \gamma + \beta \gamma)^2 - (\alpha + \beta + \gamma)(2)(\alpha \beta \gamma)$$

Where you know all the values on the RHS, so you can solve for the LHS.

Answer 3

We have \begin{eqnarray*} \alpha+\beta+\gamma =\frac{5}{2} \\ \alpha\beta+\beta\gamma+\gamma\alpha =2 \\ \alpha\beta\gamma =-3 \end{eqnarray*} Now calculate \begin{eqnarray*} (\alpha+\beta+\gamma)^2 =\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha) \\ (\alpha\beta+\beta\gamma+\gamma\alpha )^2=\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 +2 \alpha\beta\gamma(\alpha+\beta+\gamma )\\ \end{eqnarray*} So $\alpha^2+\beta^2+\gamma^2=?$ and $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=?$. Now \begin{eqnarray*} \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} =\frac{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2}{\alpha^2\beta^2\gamma^2} =\frac{?}{?} \\ \frac{1}{\alpha^2\beta^2}+\frac{1}{\beta^2\gamma^2}+\frac{1}{\gamma^2\alpha^2}=\frac{\alpha^2+\beta^2+\gamma^2}{\alpha^2\beta^2\gamma^2} =\frac{?}{?} \\ \frac{1}{\alpha\beta\gamma^2} =\frac{?}{?} \end{eqnarray*} so ...

Alternatively rearrange the original equation to $2x^3+4x=5x^2-6$. now square both sides and rearrange \begin{eqnarray*} 4x^6+16x^4+16x^2=25x^4-60x^2+36 \\ 4x^6-9x^4+76x^2-36=0 \\ 4-9x^{-2}+76 x^{-4}-36x^{-6}=0 \end{eqnarray*} So $ \color{red}{36y^3-76y^2+9y-4=0}$.