# A level Further Pure Maths help (Polynomials)

1. Sep 19, 2010

### usman94

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

Find the values of Σ(a^2), Σ(1/a), Σ(a^2)(B^2) and ΣaB(a + B) for: x^4 - x^3 + 2x + 3 = 0

2. Relevant equations

Σa = 1, ΣaB = 0, ΣaBC = -2, aBCD = 3

3. The attempt at a solution

I found the Σ(a^2) and Σ(1/a) successfully correct bt could neither find Σ(a^2)(B^2) nor (Σa)(ΣaB):
'ΣaB(a + B) = 6' is given as the answer but my own answer comes 4. Please somebody post the complete solution for ΣaB(a + B) explaining each step, given that i found ΣaB(a + B) = (Σa)(ΣaB) - 2(ΣaBc).
It seems i should have got ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc) in order to get the correct answer '6' instead of the erroneous '4'.
If perhaps this is true, then prove that ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc)

I found Σ(a^2)(B^2) = (ΣaB)^2 - 2Σ(a.B^2.C) - 4aBCD. Now 4m here i can't proceed forward to find Σ(a.B^2.C). Perhaps, i hav done it wrong or there exists an alternative easier way.
NB: a,B,C,D represent the roots alpha, beta, gamma and the 4th root (partial derivative sign) respectively.

2. Sep 19, 2010

### Staff: Mentor

I don't understand the question. What do the summations have to do with the polynomial equation? That is, how are a and B connected with x^4 - x^3 + 2x + 3 = 0?

3. Sep 19, 2010

### usman94

ohhh m terribly sorry missing to mention dat. a, B, C and D are the 4 roots of the equation: x^4 - x^3 + 2x + 3 = 0

4. Sep 19, 2010

### Staff: Mentor

OK, then what do Σ(a^2) and the other summations mean in the context of this problem?

I'm still clueless as to what this problem is asking for.

5. Sep 19, 2010

### usman94

6. Sep 19, 2010

### usman94

these summations are of the roots e.g. Σa means a + B + C + D. Likewise ΣaB represents aB + BC + CD + DA

7. Sep 19, 2010

### Staff: Mentor

The first look opens p. 294 of this book. In a quick search I didn't find exercise 9d Q2(d). The attachment in your other thread for this problem doesn't give me enough context to know what the summations are adding.

8. Sep 19, 2010

Are you working on functions of the roots of a polynomial (theory of equations?)

9. Sep 19, 2010

### epenguin

It may be helpful to write the thing out in full and then you will see there are three not two products giving you $$\alpha\beta\gamma$$

I think you are agreeing with me (?) that $$\Sigma\alpha\beta(\alpha + \beta)$$ is just $$2\Sigma\alpha^2\beta$$

You can get that out of $$\Sigma\alpha\beta\Sigma\alpha$$.

$$\Sigma\alpha\beta\Sigma\alpha$$

= 0 because

$$\Sigma\alpha\beta = 0$$.

On the other hand $$\Sigma\alpha\beta\Sigma\alpha$$ also equals in full

$$(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)(\alpha + \beta + \gamma + \delta)$$

and there are two products $$\alpha^2\beta$$ and three products $$\alpha\beta\gamma$$

$$0 = 2\alpha^2\beta - 3\alpha\beta\gamma$$

At first you find them. Later you think you need to look in the first bracket only at the ones without $$\delta$$. There are three of them. And they each have their partner in the second bracket to make $$\alpha\beta\gamma$$. Later you realise you don't need to look at each of the three because of the symmetry. And you realise there not just are but must be three - and that the $$\Sigma$$ notation is quite nifty - however you can always fall back on the full formulae if out of practice or something not working.

(I tried to bring them out by colouring the three pairs but it is already rubbish enough to try and write ordinary tex because of defects I mentioned earlier, nearly impossible to edit :grumpy:, however when you write on paper you can use colour underlining if it helps.)

Last edited: Sep 20, 2010
10. Sep 21, 2010

### epenguin

Edit: That last line should be of course

0 = 2Σα2β - 3Σαβγ

Last edited: Sep 21, 2010