Write a polynomial function of least degree with integral coefficients whose zeros include 4 and 2i ?

Question

a f(x)=x^2+4

b f(x)=x^3-4x^2+4x+16

c f(x)=x^3+4x^2+4x+16

d f(x)=x^3-4x^2-4x+16

Answer 1

If a zero is 4, then a factor is:

(x - 4)

If a zero is 2i, then there also has to be a -2i, so two more factors are:

(x + 2i)(x - 2i)

Multiply them together to get your function, since this is it in factored form:

(x - 4)(x + 2i)(x - 2i)

(x - 4)(x² - 4i²)

(x - 4)(x² + 4)

x³ + 4x - 4x² - 16

x³ - 4x² + 4x - 16

none of your list of choices.

Answer 2

None of the above. (x-4)*(x^2+4)=

x^3 - 4*x^2 + 4*x - 16

Answer 3

if 4 is a zero, then (x - 4) is a factor

if 2i is a zero, then so is -2i (because with integral coefficients, complex zeros must appear in conjugate pairs), and two more factors are (x - 2i) and (x + 2i)

factors: f(x) = (x - 4)(x + 2i)(x - 2i)

f(x) = (x - 4)(x^2 + 4)

f(x) = x^3 - 4x^2 + 4x - 16

is there a typo in one of your answers?

Answer 4

close one i would say is for D cause when i put that answer on my answer when i was in high school and i look at my test right now i got that answer correct