Question

Find the complex zeros of the following: x^3 + 216

Write in factored form.

Answer 1

Does this equation equal anything?

Answer 2

no

Answer 3

Refresh my memory. A "complex zero" is of the form a +bi, which makes the given equation =0, correct?

Answer 4

yes

Answer 5

216 is 6 cubed. So the zeros are going to be six times the zeros of x^3 + 1=0. Can you do that problem?

Answer 6

Since you want the zeros, we should set the equation equal to 0 and solve by factoring.
we get:
\[x^{3}+216 = (x+6)(x^{2}-6x+36)=0\]
Now we set both of those terms equal to 0, but the left term isnt going to produce a complex root:
\[x^{2}-6x+36 = 0 \Rightarrow x = \frac{6\pm \sqrt{36-4(36)}}{2} = 3 \pm 3i \sqrt{3}\]
If you want me to show more work let me know :)

Answer 7

(x+6)(x^2-6x+36)=0