# Finding a third-degree polynomial equation

• Jul 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?
• Jul 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 $\displaystyle z$ a complex number as a root, then its conjugate $\displaystyle \bar z$ 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 :

$\displaystyle m(x-a)(x-b)(x-c)$
• Jul 28th 2008, 10:15 AM
lax600
I think I understand the first part... but I don't understand the formula
• Jul 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 ($\displaystyle \neq 0$)
• Jul 28th 2008, 10:48 AM
lax600
I got it now thank you