Question 670349
To find the zeros of this polynomial function we need to set {{{f(x)}}} equal to {{{0}}} and factor the equation.

....since you are given  one zero which is {{{-(2i)}}}, it is result of the square of negative number that could be only {{{sqrt(-4)}}}; only {{{sqrt(-4)}}} gives you root {{{-2i}}}, as {{{2i}}} as well

therefore, you are actually given two roots, and now we will find remaining two

{{{0= x^4-32x^2-144}}}.....replace {{{-32x^2}}} with {{{-36x^2+4x^2}}}

{{{x^4-36^2+4x^2-144}}}.....group

{{{(x^4-36^2)+(4x^2-144)}}}...factor out  {{{x^2}}} from the first group and {{{4}}} from the second group

{{{x^2(x^2-36)+4(x^2-36)}}}

{{{(x^2+4)(x^2-36)}}}

{{{(x^2+4)(x-6)(x+6)}}}


Now we set each factor to zero:

{{{x^2+4=0}}}...............{{{x^2=-4}}} .........=> given roots: the imaginary numbers {{{-2i}}} and {{{2i}}}

{{{x-6=0}}}........{{{x=6}}}..................real root

{{{x+6=0}}}........{{{x=-6}}}..............real root


show it on a graph:


{{{ graph( 600, 600, -10,10, -420, 200,(x^2+4)(x-6)(x+6)) }}}