Question

Find the polynomial of lowest degree having a leading coefficient of 1, real coefficients with a zero of 6(multiplicity 2) and zero 3-2i

Answer 1

@dude @Vocaloid

Answer 2

mmm i tried but i got confused

@Shadow @jhonyy9

Answer 3

in general, if a polynomial has some zero "A", then (x-A) = 0 is one of the factors

- zero of 6 (multiplicity 2) ---> (x-6), except since it says multiplicity of 2, this zero is present twice, so this ends up being (x-6)(x-6) = 0
- zero 3-2i ---> same logic, (x - [3-2i]) = 0
- lowest degree ---> they want the lowest possible exponent on the highest degree term. we don't really have to do anything since our polynomial should have the lowest degree by default
- leading coefficient of 1 ---> by default our polynomial will have leading coefficient of 1, so we don't have to do anything here either

Answer 4

end result ---> (x-6)(x-6)(x - [3-2i]) ----> expand this, be careful with signs and the i term

Answer 5

x^4-6x^3+22x^2-30x+13?

@Mercury

Answer 6

there's only 3 x's so you shouldn't be getting an x^4 term anywhere
plus you're missing your i terms

Answer 7

so than you check it (x-6)(x-6)=(x-6)^2

yes ?

Answer 8

hope you now formula

Answer 9

know

Answer 10

@Mercury I viewed other examples of the same question, there are no i terms in the final answer.

Answer 11

@Mercury do you want me to show the work?

Answer 12

yeah nvm you're right, if 3-2i then 3 + 2i is a root