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

##### 2 Answers

2+3i, and 3.

#### Explanation:

I polynomial equations of order 3 (cubics like this) written in the form

Now because

We know

#### Explanation:

#"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#