Question 240964
<pre>
{{{f(x)=x^3-3x^2-5x+39}}}


-3 | 1 -3 -5  39
   |   -3 18 -39
    ------------
     1 -6 13   0

So now you have factored the 
polynomial f(x) this far:

{{{f(x)=(x+3)(x^2-6x+13)}}}

The zeros are found by setting
this equal to 0:

{{{(x+3)(x^2-6x+13)=0}}}

{{{x+3=0}}} give the zero -3 which was given.

{{{x^2-6x+13=0}}} will give the other zeros.

But this will not factor so we must use the
quadratic formula:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

with {{{a=1}}}, {{{b=-6}}}, and {{{c=13}}}

{{{x = (-(-6) +- sqrt((-6)^2-4*(1)*(13) ))/(2*(1)) }}}

{{{x = (6 +- sqrt(36-52 ))/2 }}}

{{{x = (6 +- sqrt(-16))/2 }}}

{{{x = (6 +- i*sqrt(16))/2 }}}

{{{x = (6 +- i*4)/2 }}}

{{{x = (6 +- 4i)/2 }}}

{{{x = 6/2 +- 4i/2 }}}


{{{x = 3 +- 2i }}} 

The other zeros are {{{3+2i}}} and {{{3-2i}}}

Edwin</pre>