Question

Write a polynomial function of least degree with integral coefficient that has the given zeros.
6, 2+2i

Answer 1

Basically, 6 and 2 + 2i are two solutions of the polynomial function when f(x) = 0. In other words,

x = 6
x = 2 + 2i

But remember that solutions involving complex numbers occur as conjugate pairs so

x = 2 - 2i is also a solution.

Now next, we can move all the numbers to one side so that each equation is equal to zero:

x - 6 = 0
x - 2 - 2i = 0
x - 2 + 2i = 0

And next, we can reference the zero product property to show that

(x - 6)(x - 2 - 2i)(x - 2 + 2i) = 0

From here, we can multiply the left hand side and we will have the polynomial of simplest form that includes all the given zeroes.

Answer 2

So if you multiply:

(x - 6)(x - 2 - 2i)(x - 2 + 2i) you will get the polynomial function of least degree.

Answer 3

@DSoul_64 does any of what I have shown so far make any sense to you at all?

Answer 4

A little

Answer 5

Where exactly are you confused? Would you mind explaining what you understand, or more importantly what you do not understand?

Answer 6

Mostly when to multiply the (x-2+2i) and (x-2-2i)

Answer 7

Be back in 30 min

Answer 8

Suppose we were working with real numbers only. The question states that the zeroes of a polynomial are 2, 3, and 4.

You know that this means the solutions of the polynomial where f(x) = 0 occur where

x = 2
x = 3
x = 4

You also know that if you move everything to one side you will have

x - 2 = 0
x - 3 = 0
x - 4 = 0

Now, to get the polynomial that includes these x values, you have to write out this equation in reference to the zero product property:
(x - 2)(x - 3)(x - 4) = 0

Then multiply the binomials on the left side to get:

x^3 - 9x^2 + 26x - 24 = f(x)