What are the zeros for #x^ { 3} + 9x ^ { 2} + 4x + 36#?

1 Answer
Nov 9, 2017

The roots are #+ 2i, -2i, -9#.

Explanation:

We can factor as following:

#x^2(x + 9) + 4(x + 9)#

#(x^2 +4)(x + 9)#

If we set to #0#, we get:

#(x^2 + 4)(x +9) = 0#

Solving for #x#, we get #x = -9# or #x^2 = -4#. So the imaginary roots will be #x = +-sqrt(-4) = +- 2i#.

Hopefully this helps!