Question

A 4th-degree polynomial function has zeros at 3 and 5-i. Can 4+i also be a zero of the function? Please explain how.

Answer 1

hint
if \(a+bi\) s a zero of a real polynomial, what else has to be a zero?

Answer 2

I'm not too sure, but maybe \[a-bi\]

Answer 3

yes

Answer 4

so if \(5-i\) is a zero, what else has to be a zero?

Answer 5

\[5+i\] ?

Answer 6

right, so now you have three zeros
\[3,5+i,5-i\] how many zeros can a 4th degree polynomial have?

Answer 7

A 4th-degree polynomial can have 4 zeros.

Answer 8

yea
so can \(4+i\) be a zero?

Answer 9

well actually it could, but not if the polynomial has REAL coefficients, which is probably what is meant here

Answer 10

ok yeah, but if the coefficients are real, then if \(4+i\) is a zero, so is \(4-i\) and now they are too many

Answer 11

lol "there are too many"

Answer 12

Wait, do polynomial functions with complex coefficients have at least one complex zero?

Answer 13

fundamental theorem o' algebra says any polynomial has at least one zero

Answer 14

doesn't say if the zeros are real or complex though...

Answer 15

which means, counting multiplicity, a polynomial of degree \(n\) has \(n\) zeros

Answer 16

Okay, so I looked up the theorem, and a statement said: "Every polynomial function of degree \[n \ge 1\] has at least one complex zero."

Answer 17

true that, as the kids say

Answer 18

In theory, polynomial functions with complex coefficients do have at least one complex zero; as noted in complex polynomial function, f(x), of a degree \( \sf n \ge \) 1.

Answer 19

add that to the "factor theorem" and you get what @satellite73 said

Answer 20

But I still don't understand. The polynomial function has real coefficients, so 4+i cannot be a zero because...?

Answer 21

this still doesn't really answer the question
i am going to guess that whoever wrote it, had in mind polynomials with real coefficients
you are not at the point to understand what a function with complex coefficients would mean, although i could be wrong

Answer 22

beause so would \(4-i\) so you would have 5 zeros \[3,5+i,5-i,4+i,4-i\] one too many

Answer 23

Though, also note that a polynomial, \( f\), of an odd degree with \( \sf real~coefficients\) will have at least one real root

Answer 24

Thank you, @satellite73, and everyone else that helped me understand this better! So, because 5-i is a zero, 5+i is a zero. Then, 3 is a zero, along with another unknown zero. That zero cannot be 4+i because, if 4+i were a zero, then there would be a 4-i, which is one too many.

Answer 25

Theorem states there has to be at least be as many roots as the highest degree. I'd inferring looking into the Conjugate Pair Theorem: r = a - bi is a further zero of \( f \)

( I couldn't find the LaTex for 'r')