Polynomial Functions

Question

Find a polynomial function that has the zeros.

A. 0,3, +2i B. +2i, + 3i, 4 C. -2, 3, 3i D. -i, 2i, -3i

E. -4, 1+ 2i F. 1-i, 1+3i, 0

Answer 1

Hello Joann

To do these kinds of problems.

First you need to know what is the definition of zeros means.

Zero means the value of x that make the polynomial equation equal to 0.

Therefore, when you factor a polynomial out the zeros will be part of the factor.

Ex:

1. Find polynomial that have zeros of 0, 3, +2i.

Now since there is zero of 0. It means when you factor the polynomial it will have this form (x - 0)

zero of 3, there will also a factor in this form ( x-3)

zero of +2i.

Special case of imaginary zeros

For this type of problem you have to know that i means square root of -1

Therefore, if zeros value is imaginary number you always have another pair of imaginary value.

Ex, zero of polynomial is +2i, another imaginary zeros will be -2i

Therefore the polynomial contain the following factors.

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

Now after you found all the factors, next step is multiply them together

(x- 0) (x-3) (x-2i ) (x+2i) and you will get the polynomial.

x (x-3) (x +4)

Let me know if you need more help.

Answer 2

Hi Joann, the easiest way to do this is to define your function in terms of factors. So, for part A for example, if you define your function as f(x) = (x-0)(x-3)(x-2i)(x-(-2i)), then f has zeroes at 0, 3, 2i, and -2i. To get your function in standard form, you just need to multiply the factors. It's usually easiest to multiply the factors with complex roots first, since some of the i's will multiply to -1 and you'll end up with some real numbers:

(x-0)(x-3)(x-2i)(x-(-2i))

= (x-0)(x-3)(x-2i)(x+2i)

= (x-0)(x-3)(x^2+2ix-2ix-4i^2)

= (x-0)(x-3)(x^2-4i^2)

= (x-0)(x-3)(x^2-4(-1))

= (x-0)(x-3)(x^2+4)

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

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

= x^4-3x^3+4x^2-12x

You can use the exact same process with the rest of the parts, hope this helps!