Finding a third-degree polynomial equation

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July 28th 2008, 09:55 AM
lax600

Finding a third-degree polynomial equation

I have to find a third-degree polynomial equation with rational coefficients that has the given numbers as roots.

1 and 3i

I have no idea how to do this, and it's due tomorrow. I've checked my textbook, online but can not find an explanation. Could someone help me please?
July 28th 2008, 10:02 AM
Moo

Hello !

Quote:

Originally Posted by lax600

I have to find a third-degree polynomial equation with rational coefficients that has the given numbers as roots.

1 and 3i

I have no idea how to do this, and it's due tomorrow. I've checked my textbook, online but can not find an explanation. Could someone help me please?

There are 2 things that will be useful here :
- if a is a root of a polynomial, then this polynomial can be factored by (x-a)
- if a polynomial has rational coefficients, and $z$ a complex number as a root, then its conjugate $\bar z$ is also a root.

So 1 is a root. Since it's a polynomial with rational coefficients, 3i is a root, but -3i too.
You've got the 3 roots, so now the polynomial will be in the form :

$m(x-a)(x-b)(x-c)$
July 28th 2008, 10:15 AM
lax600

I think I understand the first part... but I don't understand the formula
July 28th 2008, 10:19 AM
Moo

Quote:

Originally Posted by lax600

I think I understand the first part... but I don't understand the formula

Ok, a,b and c are the roots.

But there's still a factor (m) that doesn't change the roots of the polynomial. Why ? Because a root a of a polynomial P is such that P(a)=0. So it doesn't matter if you multiply the polynomial by a number ( $\neq 0$ )
July 28th 2008, 10:48 AM
lax600

I got it now thank you