Question

HELP PLEASE!!
Write the polynomial that has the given numbers as roots. (Picture below)

Answer 1

Assuming the polynomial you are looking for has real coefficients, then the conjugate of each of the complex roots is also a root. Therefore you are looking for a fifth degree polynomial with real coefficients, and roots are {5, 3+2i, 3-2i, 1-3i, 1+3i}.

Since a monic polynomial with roots {x1,x2,...xn) is formed by:
f(x)=(x-x1)(x-x2)...(x-xn)
the polynomial (with real coefficients) you are look for has form:
f(x)=(x-5)(x-(3+2i))(x-(3-2i))(x-(1-3i))(x-(1+3i))
All you need to do is to expand the expression.

When expanding, it is easier arithmetically to expand the conjugates by pairs, thereby eliminating the complex terms at the first steps.

Answer 2

Oh okay, I see! Thank you so much :)

Answer 3

You're welcome! :)

Answer 4

Oh wait, quick question @mathmate ! If I was expanding the 3+2i and 3-2i (using the FOIL method), then how would 2i X -2i work?

Answer 5

oh, would it just be 2? Since i is equal to -1

Answer 6

With (3+2i)(3-2i), I would not use FOIL, I would use "difference of squares".
(3+2i)(3-2i)=(3)^2-(2i)^2=9-(2^2)(i^2)=9-4=5
However, you are actually going to expand:
(x-(3+2i))(x-(3-2i))=(x-3 - 2i)(x-3 + 2i)
you can work with the difference of squares in a similar way.

Answer 7

Okay, ahh! Haha, but I see where you are going :)
Thanks again!!

Answer 8

Correction:
*...9-(2^2)(i^2)=9\(\color{red}{+}\)4=\(\color{red}{13}\)