# Homework Help: 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:

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

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