# Coefficients of polinom

1. Mar 10, 2008

### Theofilius

1. The problem statement, all variables and given/known data

What should the coefficients a,b,c of the polinom $$P(x)=x^3+ax^2+bx+c$$ be, so his roots $$x_1+x_2=x_3$$ ?

2. Relevant equations

$$P(x)=a_n(x-c_1)(x-c_2)...(x-c_n_-_1)(x-c_n)$$

$$c_1+c_2+...+c_n= -\frac{a_n_-_1}{a_n}$$

$$c_1c_2+c_2c_3+...+c_n_-_1c_n= \frac{a_n_-_2}{a_n}$$

$$c_1c_2c_3+c_1c_2c_4+...+c_n_-_2c_n_-_1c_n=-\frac{a_n_-_3}{a_n}$$

...........................................................................................

$$c_1c_2...c_n_-_1 + c_1c_2...c_n_-_2c_n+......+c_2c_3...c_n=(-1)^n^-^1 \frac{a_1}{a_n}$$

$$c_1c_2c_3...c_n= (-1)^n \frac{a_0}{a_n}$$

3. The attempt at a solution

I don't know where to start from. Anybody have any idea? Thnx for the help.

Last edited: Mar 10, 2008
2. Mar 10, 2008

### SlideMan

I think you forgot parts 2 and 3. ;)

3. Mar 10, 2008

### rock.freak667

Normally you should show some working or at least some idea of what you think you must do. But if $P(x)=x^3+ax^2+bx+c$, I can think of many cubic polynomials in which the roots will be $\alpha,\beta,\alpha+\beta$. I am assuming that you are to find a cubic polynomial with leading coefficient 1 and has roots $\alpha,\beta,\alpha+\beta$.

For a cubic polynomial of the form $ax^3+bx^2+cx+d=0$
$$\sum \alpha = \frac{-b}{a}$$

$$\sum \alpha\beta = \frac{c}{a}$$

$$\sum \alpha\beta\gamma =\frac{d}{a}$$

(Note: $\sum \alpha$ denotes the sum of the roots taking one at a time)

4. Mar 10, 2008

### Theofilius

Ok... Sorry... Is it ok, now?

5. Mar 10, 2008

### rock.freak667

Yes that is how to start but as I was saying before, there are many polynomials whose roots can be $\alpha,\beta,\alpha+\beta$. Try solving for the roots to be in terms of the coefficients of the polynomial.

6. Mar 10, 2008

### Theofilius

Can you please start just a little bit, with solving, so I can go on?

7. Mar 10, 2008

### rock.freak667

$$\sum \alpha=\frac{-b}{a}$$

For P(x),a=1 and b=a

so that

$$\sum \alpha=\frac{-a}{1}=-a$$

$$\sum \alpha =\alpha+\beta+\gamma=-a$$

and you know that $\gamma=\alpha+\beta[/tex] so for [itex]\sum \alpha$ you really get

$$2\gamma =-a$$

$$\gamma=\frac{-a}{2}$$

Can you go on from here?

8. Mar 10, 2008

### Theofilius

Ok.
$$x_1+x_2+x_3=-a$$

$$x_1+x_1x_3+x_2x_3=b$$

$$x_1x_2x_3=-c$$

------------------------------------

$$x_3+x_3=a$$

$$x_3(x_1+x_2)+(x_3-x_2)x_2=b$$

$$(x_3-x_2)x_3=-c$$

----------------------------------------

$$2x_3=a$$

$$x^2_3+x_3x_2-x^2_2=b$$

$$x^2_3-x_2x_3=-c$$

I am stuck here. Can you help me please?

9. Mar 10, 2008

### tiny-tim

Theofilius, you haven't used x1 + x2 + x3 …

I believe in trying to find an easy way.

We know P(x) only has three roots.

So P(x) = (x+p)(x+q)(x+r), with r = p + q.

So P(x) = (x+p)(x+q)(x+p+q).

So a = 2p + 2q, b = … , c = … ?

10. Mar 10, 2008

### Theofilius

Don't know how to continue... Can you please give some hint more :D?

11. Mar 10, 2008

### tiny-tim

P(x) = (x+p)(x+q)(x+p+q)

So P(x) =
x^3
+ x^2(p + q + (p + q))
+ x(pq + p(p+q) + q(p+q))
+ pq(p+q)​

But we are given that P(x) = x^3 + ax^2 + bx + c.

So the two expressions must be equal.

So a = 2p + 2q, b = … , c = … ?

12. Mar 10, 2008

### Theofilius

$$a=2p+2q, b=p^2+q^2+3pq , c=p+q$$

What is next?

13. Mar 10, 2008

### tiny-tim

Careful!

It's actually $$a=2p+2q, b=p^2+q^2+3pq , c=pq(p+q)\,.$$

And it's simpler if you leave the middle one as $$b=(p+q)^2+pq\,.$$

Anyway, next you divide c by a, giving you pq = 2c/a;

and then you fiddle around with b …

14. Mar 10, 2008

### Theofilius

Can you give me just the solution, so I will understand what you do. Like this part by part, I can't "pack" the parts.

15. Mar 10, 2008

### tiny-tim

Theofilius, I haven't seen any sign that you do understand what I do.

It's no good understanding what other people do unless you can do it yourself.

This is a very simple problem, but you're stiil having difficulty with it.

You need the practice!

16. Mar 11, 2008

### Theofilius

Ok, how will I practice, when I don't know how to solve this problem. Can you please tell me the principle of solving, so I can do it. What's next?

17. Mar 11, 2008

### Theofilius

Help!

18. Mar 12, 2008

### Theofilius

I came up with this system.

$$2x_3 = -a$$
$$x_3^2 - \frac{c}{x_3}=b$$
$$x_1x_2 = - \frac{2c}{a}$$

I sow in my text book results, and the result is:

$$a^3 - 4ab + 8c = 0$$

19. Mar 12, 2008

### tiny-tim

Yes, that looks fine.

Your only problem is you've lost track of what your target is.

Your target is to find an equation involving a b and c, but not x1 x2 or x3.

Can you see how to use the three equations above to get you an equation involving only a b and c?

(Try … if you can't do it, I'll help you … but it's actually very easy, so take some time before giving up …)

20. Mar 12, 2008

### Physicsissuef

Substitute for [itex]x_3[/tex] from the 1st into the 2nd equation, like this:

$$x_3 = - \frac{a}{2}$$

$$\frac{a^2}{4} + \frac{2c}{a} - b =0$$

Now multiply by 4a, and problem solved.