Polynomial coefficient problem

Hello, I have an another unsolved problem from Gelfand's book on Algebra, here it is:

Problem 153. The polynomial has three different roots (the authors guarantee it) denoted by . Write a polynomial with integer coefficients having roots

(a) ; (b) ; (c)

I understand that , but that's pretty much all I can 'do', hints would be appreciated, thanks in advance.

Re: Polynomial coefficient problem

Hello, PaulAdrienMauriceDirac!

I'll do part (a).

I hope you understand the reasoning.

We know that: .

Sum of roots: .

Sum of pairs: .

. . . . . . . . .

. . . . . . . . .

Product of roots: .

. . . . . . . . . . .

. . . . . . . . . . .

Therefore: .

Re: Polynomial coefficient problem

Thank you very much! I didn't know that Vieta's theorem existed for cubic equations, your explanation is very clear.

Re: Polynomial coefficient problem

Hi PaulAdrienMauriceDirac,

P(x) = (x - x1)(x - x2)(x - x3) = x^3 - (x1 + x2 + x3)x^2 + (x1x2 + x1x3 + x2x3)x - x1x2x3.

So, equating co-efficients like this holds for cubics, and in general, all polynomial equations roots.

Salahuddin

Maths online

Re: Polynomial coefficient problem

While Vieta's theorem is the way to go for parts b) and c), for part a) you could simply shift the polynomial 1 unit to the right to add 1 to the roots:

Re: Polynomial coefficient problem

Quote:

Originally Posted by

**MarkFL2**

Oh, that's so much easier, too bad I haven't thought about it, thank you very much!

Re: Polynomial coefficient problem

The same sort of method will work for part (b) as well, replace x by x/2.

The resulting polynomial is

Re: Polynomial coefficient problem

Re: Polynomial coefficient problem

Your polynomial is simply eight times mine, they will have the same zeros.