# Roots of polynomial equations

• Dec 26th 2010, 01:38 PM
Femto
Roots of polynomial equations
I was happily answering some questions from my further pure textbook earlier until I encountered the last question; forgive me in advance for my lack of understanding, I'm only a sixth form student at the moment and I don't have as much knowledge as most people on this site!

Here's the question:

The roots of the equation $x^3 + ax + b$ are $\alpha, \beta, \gamma$. Find the equation with roots $\frac{\beta}{\gamma} + \frac{\gamma}{\beta}, \frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}, \frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.

Initially I tried to use substitution but that failed epicly. Overall, I'm really confused - please could somebody assist me?

Also, on a side note, I'm new here! Just thought you'd like to know (Smile)
• Dec 26th 2010, 01:44 PM
dwsmith
Since $\alpha, \ \beta, \ \mbox{and} \ \gamma$ are roots, $x^3+ax+b=(x-\alpha)(x-\beta)(x-\gamma)$

Does this help?
• Dec 26th 2010, 01:46 PM
Femto
Quote:

Originally Posted by dwsmith
Since $\alpha, \ \beta, \ \mbox{and} \ \gamma$ are roots, $x^3+ax+b=(x-\alpha)(x-\beta)(x-\gamma)$

Does this help?

Yes I kind of understand, but how do I obtain another equation with the new fractional roots presented in the question?

Would it perhaps make sense to write it out as this?

$(x - (\frac{\beta}{\gamma} + \frac{\gamma}{\beta}))(x - (\frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}))(x - (\frac{\alpha}{\beta} + \frac{\beta}{\alpha}))$

Really sorry, I'm getting a tad confused.
• Dec 26th 2010, 01:49 PM
dwsmith
$\displaystyle \left(x-\left(\frac{\beta}{\gamma}+\frac{\gamma}{\beta}\ri ght)\right)(\cdots)(\cdots)$
• Dec 26th 2010, 01:51 PM
dwsmith
Quote:

Originally Posted by Femto
Yes I kind of understand, but how do I obtain the new equation with the new fractional roots presented in the question?

Would it help to say:

$(x - (\frac{\beta}{\gamma} + \frac{\gamma}{\beta}))(x - (\frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}))(x - (\frac{\alpha}{\beta} + \frac{\beta}{\alpha}))$?

Now multiply it out.

I would leave it factored, but if you need it in x^3.... format, you need to multiply.
• Dec 26th 2010, 01:56 PM
Femto
Quote:

Originally Posted by dwsmith
Now multiply it out.

I would leave it factored, but if you need it in x^3.... format, you need to multiply.

Thanks; yikes this looks time consuming.
• Dec 26th 2010, 01:58 PM
dwsmith
$x^3+ax+b=(x-\alpha)(x-\beta)(x-\gamma)=x^3-\alpha x^2-\gamma x^2-\beta x^2+\alpha\beta x+\beta\gamma x+\alpha\gamma x-\alpha\beta\gamma$

$b=-\alpha\beta\gamma$

$ax=x\alpha\beta +x\beta\gamma+x\alpha\gamma$

$0x^2=-\alpha x^2-\gamma x^2-\beta x^2$
• Dec 26th 2010, 02:04 PM
Plato
Quote:

Originally Posted by Femto
Here's the question:
The roots of the equation $x^3 + ax + b$ are $\alpha, \beta, \gamma$. Find the equation with roots $\frac{\beta}{\gamma} + \frac{\gamma}{\beta}, \frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}, \frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.

From the given we know that $\alpha+\beta+\gamma=0$,
$\alpha\beta+\alpha\gamma+\beta\gamma=a$ and $\alpha\beta\gamma=-b$.
You can use those and multiply out what you have setup.