Polynomial roots

1. Jul 6, 2009

Mentallic

1. The problem statement, all variables and given/known data
Given a general cubic $$a_1x^3+b_1x^2+c_1x+d_1=0$$ has roots $$\alpha,\beta,\gamma$$

find the polynomial $$a_2x^3+b_2x^2+c_2x+d_2=0$$ that has roots $$\alpha ^2,\beta ^2,\gamma ^2$$

3. The attempt at a solution
$$\alpha ^2+\beta ^2+\gamma ^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma)$$

Thus, $$\alpha ^2+\beta ^2+\gamma ^2=(\frac{-b_1}{a_1})^2-2(\frac{c_1}{a_1})=\frac{b_1^2-2a_1c_1}{a_1^2}$$

Therefore, $$-\frac{b_2}{a_2}\equiv -\frac{b_1^2-2a_1c_1}{a_1^2}$$

So the new polynomial is now in the form:

$$a_1^2x^3+(2a_1c_1-b_1^2)x^2+c_2x+d_2=0$$

Also, $$\alpha ^2\beta ^2\gamma ^2=(\alpha\beta\gamma)^2$$

Thus, $$\alpha ^2\beta ^2\gamma ^2=(\frac{d_1}{a_1})^2=\frac{d_1^2}{a_1^2}$$

Therefore, $$-\frac{d_2}{a_2}\equiv -\frac{d_1^2}{a_1^2}$$

So now the polynomial is:

$$a_1^2x^3+(2a_1c_1-b_1^2)x^2+c_2x+d_1^2=0$$

In order to find $$c_2$$ in terms of the coefficients of the first polynomial, I'll need to express
$$\alpha ^2\beta ^2+\alpha ^2\gamma ^2+\beta ^2\gamma ^2$$ in terms of sum of roots one, two and three at a time, using the similar idea as was done to find the sum of the squared roots one at a time. However, I'm unsure how to do this. Please help.

2. Jul 6, 2009

rock.freak667

You'd need to consider

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

3. Jul 9, 2009

Count Iblis

You can also substitute x = sqrt(y) in the equation. Then that immediately guarantees that the roots in terms of y are the squares of the roots of the original equation. To get rid of the fractional powers of y, you bring the y^(1/2) and y^(3/2) terms to one side of the equation and then you square both sides.

4. Jul 10, 2009

Mentallic

This question was in an exam I had:

Yes I had first considered to substitute x=sqrt(y) but after realizing it wasn't a polynomial anymore and from what our teacher said previously in class "if it isn't a polynomial then it wont work". I didn't consider squaring to form it into a 3rd degree polynomial again, basically because my trust in it working wasn't there after what my teacher had said.

This is the approach I took to find the roots, but I'm unsure what to do with the long expansion:

$$(\alpha ^2+\beta ^2+\gamma ^2)^2=\alpha ^4+\beta ^4+\gamma ^4-2(\alpha ^2\beta ^2+\alpha ^2\gamma ^2+\beta ^2\gamma ^2)$$

As for the 3 roots to the 4th degree, is it really going to be that complicated to solve this problem in this way?

5. Jul 10, 2009

Count Iblis

With these sorts of problems there is usually always a straightforward brute force method available which often involves solving eqations to get things exactly right. But if you think carefully, you can often find a method that yields the answer without much effort.

See here another such problem:

https://www.physicsforums.com/archive/index.php/t-8259.htmlbr/t-263816.html

6. Jul 10, 2009

Mentallic

If you're referring to the post you made in that thread, indeed your solution wasn't the conventional brute force method, but the problem is that I wouldn't have been able to think carefully in that way.

7. Jul 11, 2009

Дьявол

Ok, you missed this one:

$$\alpha\beta +\alpha\gamma+\beta\gamma=\frac{c_1}{a_1}$$

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

Can you continue?

Regards.

8. Jul 11, 2009

Mentallic

Oh jeez that is MUCH more simple! I guess rock.freak667 and I went off and chose a much more complicated expansion, to our doom Thanks Дьявол

ok so basically just to finish it off:

$$\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2=(\alpha\beta +\alpha\gamma+\beta\gamma)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)=(\frac{c_1}{a_1})^2-2(\frac{-d_1}{a_1})(\frac{-b_1}{a_1})=\frac{c_1^2-2b_1d_1}{a_1^2}$$

Therefore the polynomial is:

$$a_1^2x^3+(2a_1c_1-b_1^2)x^2+\frac{c_1^2-2b_1d_1}{a_1^2}x+d_1^2=0$$