If 2-3i is a root of #z^3-7z^2+25z-39=0#, find the other two roots?

I polynomial equations of order 3 (cubics like this) written in the form #ax^3 + bx^2 + cx + d = 0#, with three roots, #alpha#, #beta#, and #gamma#, it is a rule that #alpha + beta + gamma = -b-:a#, also #alpha*beta + beta*gamma + gamma*alpha = c-:a#, and that #alpha*beta*gamma = -d-:a#.
Now because #d# is real and not imaginary, that means #alpha*beta*gamma# is real. You should know that to get a real number by multiplying imiginary numbers, you times it by it's conjugate, which for #2-3i# is #2+3i#. So now we know that #alpha# and #beta# are #2-3i# and #2+3i#.
We know #alpha*beta*gamma = -d-: a#, so #(2-3i)(2+3i)gamma = -(-39)-:1#. Expanding #(2-3i)(2+3i)# we get #13#, so #13gamma = 39#, thus #gamma = 39-:13 = 3#, therefore giving us the other two roots: #2+3i# and #3#.

#"the complex roots of polynomial equations always "#
#"occur in "color(blue)"conjugate pairs"#

#2-3i" is a root "rArr2+3i" is also a root"#

#"the quadratic factor formed by these roots is "#

#(z-(2-3i))(z-(2+3i))#

#=((z-2)+3i)((z-2)-3i)#

#=(z-2)^2-9i^2#

#=z^2-4z+4+9#

#=z^2-4z+13#

#rArrz^3-7z^2+25z-39#

#=color(red)(z)(z^2-4z+13)color(magenta)(+4z^2)-7z^2color(magenta)(-13z)+25z-39#

#=color(red)(z)(z^2-4z+13)color(red)(-3)(z^2-4z+13)color(magenta)(-12z)+12zcolor(magenta)(+39)#
#color(white)(=)-39#

#=color(red)(z)(z^2-4z+13)color(red)(-3)(z^2-4z+13)+0#

#rArr(z-3)" is a root"#

#rArr(z-3)(z-(2-3i))(z-(2+3i))=0#

#rArr"roots are " z=3" and " z=2+-3i#