Question

find the complex zeros of x^3+27. write f in factored form

Answer 1

Difference of 2 cubes
(x+3)(x^2-3x+9)=0
We know one zero is -3, you can apply the quadratic formula to x^2-3x+9 to find the 2 complex zeros

Answer 2

\[x=\frac{3 \pm \sqrt{9-4(1)(9)}}{2(1)}\]
\[x=\frac{3 \pm 3i \sqrt{3}}{2}\]

Answer 3

\[f(x)=(x+3)(x-\frac{3-3i \sqrt{3}}{2})(x-\frac{3+3i \sqrt{3}}{2})\]

Answer 4

i said difference of 2 cubes above, i meant sum of 2 cubes, but i applied the right formula.

Answer 5

thank you