Monic polynomial of the lowest possible degree

[53Mark53]

Homework Statement

A monic polynomial is a polynomial which has leading coefficient 1. Find the real, monic polynomial of the lowest possible degree which has zeros −1−2i,−2i and i. Use z as your variable.

The Attempt at a Solution

[/B]
Would I just expand the zeros giving me

(z+1+2i)(z+2i)(z-i)

z^3+(1+3*I)*z^2+I*x+I*4+2

 

[SammyS]

Homework Statement

A monic polynomial is a polynomial which has leading coefficient 1. Find the real, monic polynomial of the lowest possible degree which has zeros −1−2i,−2i and i. Use z as your variable.

The Attempt at a Solution

[/B]
Would I just expand the zeros giving me

(z+1+2i)(z+2i)(z-i)

z^3+(1+3*I)*z^2+I*x+I*4+2

Don't forget: It's a real polynomial.

 

[53Mark53]

Don't forget: It's a real polynomial.

how would I make it real?

 

[SteamKing]

how would I make it real?

Hint: what happened to the conjugates of the original roots?

 

[53Mark53]

Hint: what happened to the conjugates of the original roots?

Hint: what happened to the conjugates of the original roots?

Does that mean that I have to square the imaginary roots?

 

[SteamKing]

Does that mean that I have to square the imaginary roots?

What is the conjugate of the complex number, a + bi?

 

[SammyS]

@53Mark53 ,
Here's what you said in a previous thread:

Homework Statement

p(x) = x^3 − x^2 + ax + b is a real polynomial with 1 + i as a zero, find a and b and find all of the real zeros of p(x).

The Attempt at a Solution

[/B]
1-i is also a zero as it is the conjugate of 1+i

The same applies here, but now you have 3 complex roots, no two of which form a conjugate pair.

 

[53Mark53]

@53Mark53 ,
Here's what you said in a previous thread:

The same applies here, but now you have 3 complex roots, no two of which form a conjugate pair.

Thanks I got the right answer by using conjugate pairs