Question 919829

given zeros:

{{{-5}}}, 
{{{3+2i}}}, ...since we have {{{3+2i}}}, there must be {{{3-2i}}} too
{{{3-2i}}}
{{{-1}}}, 
{{{5}}}, 
{{{2}}} 

use zero product rule:


{{{(x+5)(x-(3+2i))(x+(3+2i))(x+1)(x-5)(x-2) }}}... multiply all,first group them and multiply each group

{{{((x+5)(x-5))((x-(3+2i))(x+(3+2i)))((x+1)(x-2)) }}}

{{{(x^2-5^2)((x-3-2i))(x+3+2i))(x^2-2x+x-2) }}}

{{{(x^2-25)(x^2+cross(3x)+cross(2xi)-cross(3x)-9-6i-cross(2xi)-6i-4i^2)(x^2-x-2) }}}

{{{(x^2-25)(x^2-x-2)(x^2-9-12i-4(-1)) }}}

{{{(x^4-x^3-2x^2-25x^2+25x+50)(x^2-9-12i+4) }}}

{{{(x^4-x^3-27x^2+25x+50)(x^2-12i-5) }}}

{{{x^6-x^5-32x^4+30x^3+185x^2+i(-12x^4+12x^3+324x^2-300x-600)-125x-250}}}