Question

Find the remaining zeros of f in a polynomial f(x) which has coefficients that are real. Degree: 3, zeros: 4, 5 - i

Answer 1

@julian25 PLEASE HELPPPPPP

Answer 2

Degree 3 means 3 solutions. You have found 3. If it was degree 4 I could help

Answer 3

its not 5, -i its 5 minus i so is the remaining zero 5 + i?

Answer 4

\[(x-4)(x^2-h) = 0\]Where h is something we don't know yet. Obviously x = 4 is a solution. and \[x^2 = h\]\[x = \pm\sqrt h\]We already know that one of the square root of h's is \(5-i\) So we can say that the other is simply the negative.So strangely I don't think it is the conjugate but simply the negative. Does that logic hold?

Answer 5

so then to negate it it would become -5+i I think. this one is confusing

Answer 6

Yeah, but that does seem odd. Might be worth re asking and getting someone who is mega sure. Although I can't see a flaw with that solution, can you?

Answer 7

@modphysnoob

Answer 8

I really dont see a flaw but I suck at math

Answer 9

4, 5-i, 5+i

Answer 10

are you mega sure?

Answer 11

yes