For each of the following complex polynomials, one of its zeroes is given. Use this zero to help write the polynomial in completely factored form.

a. C(z) = z^3+(1-4i)z^2+(-6-4i)z+24i ; z=4i

answer - C(z) = (z-4i)(z+3)(z-2)

Gotta love having the answer, but no clue how to solve.

They give you the one zero (in this degree-3 poly) because then you can divide that out and then use the

Quadratic Formula to do the rest. So do the

long poly division to take out that one zero / factor (b/c if x = a is a zero then x - a is a factor). Set it up the regular way:

Code: Select all

```
long div:
--------------------------------------
z - 4i ) z^3 + (1 - 4i)z^2 + (-6 - 4i)z + 24i
```

Then do the division the regular way:

Code: Select all

```
long div:
z^2
--------------------------------------
z - 4i ) z^3 + (1 - 4i)z^2 + (-6 - 4i)z + 24i
z^3 + ( - 4i)z^2
z^2
--------------------------------------
z - 4i ) z^3 + (1 - 4i)z^2 + (-6 - 4i)z + 24i
z^3 + ( - 4i)z^2
------------------
(1 )z^2 + (-6 - 4i)z + 24i
z^2 + 1z
--------------------------------------
z - 4i ) z^3 + (1 - 4i)z^2 + (-6 - 4i)z + 24i
z^3 + ( - 4i)z^2
------------------
(1 )z^2 + (-6 - 4i)z + 24i
(1 )z^2 + ( - 4i)z
z^2 + 1z
--------------------------------------
z - 4i ) z^3 + (1 - 4i)z^2 + (-6 - 4i)z + 24i
z^3 + ( - 4i)z^2
------------------
(1 )z^2 + (-6 - 4i)z + 24i
(1 )z^2 + ( - 4i)z
-------------------------
(-6 )z + 24i
z^2 + 1z + -6
--------------------------------------
z - 4i ) z^3 + (1 - 4i)z^2 + (-6 - 4i)z + 24i
z^3 + ( - 4i)z^2
------------------
(1 )z^2 + (-6 - 4i)z + 24i
(1 )z^2 + ( - 4i)z
-------------------------
(-6 )z + 24i
(-6 )z + 24i
```

This leaves you with something you can factor or use the Formula on. So it's what they show

here but with complexes.