Question 1113596
The zeros at 3 and -13  result in factors (x-3)(x+13)

The zero at 5+4i  will be one of two zeros, where the 2nd zero is at 5-4i  (complex roots come in conjugate pairs).   

Assuming the form {{{ x^2 + bx + c = 0 }}}  and using the quadratic formula 
{{{ -(b/2a) +- sqrt((b^2-4ac))/(2a) }}}

             {{{   -b/(2a) =  5 }}} —>  {{{ b = -10 }}}
             {{{   sqrt((b^2-4*1*c))/(2*1) = 4i }}} —>   {{{ (100-4c)/4 = -16 }}}  —>  {{{ 100-4c = -64 }}}
               —> {{{ c = 41 }}}

    {{{ x^2 - 10x + 41 }}}  generates  zeros at {{{ 5 +- 4i }}} 

—
Putting it all together:   {{{ f(x) = (x - 3)(x + 13)(x^2 - 10x + 41) }}}

…in expanded form:  {{{ highlight(f(x) = x^4 - 98x^2 + 800x - 1599) }}}