SOLUTION: Help!! Find the complex zeros of the polynomial function. Write f in factored form. f(x)=x^3-8x^2+29x-52

First we try to find any real rational zeros.

The only possible rational zeros are ± the factors of 52. They are
±1, ±2, ±4, ±13, ±26, ±52

Try 1 using synthetic division

1 | 1 -8 29 -52
  |    1 -7  22
    1 -7 22 -30   Nope. f(1)=-30, not 0.


Try -1 using synthetic division

-1 | 1 -8 29 -52
   |   -1  9 -38
     1 -9 38 -90   Nope. f(-1)=-90, not 0.


Try 4 using synthetic division

4 | 1 -8  29 -52
  |    4 -16  52
    1 -4  13   0   Yes! f(4)=0.

That means (x-4) is a factor of f(x), and the quotient is
1x²-4x+13 (the numbers left of the 0 in the synthetic division.
So we have this factorization of f(x)

f(x) = (x-4)(x²-4x+13)   <-- that's the simplest factored form without 
                             using complex numbers!

We set that = 0
       (x-4)(x²-4x+13) = 0
        
     x-4=0;  x²-4x+13 = 0
       x=4      
             
              
             
             
             
             
             
             

So f(x) has real zero 4, and complex zeros 2+3i and 2-3i.

So allowing complex numbers we can find another factored form
for f(x) of only linear factors:

f(x) = (x-4)[x-(2+3i)][x-(2-3i)]

f(x) = (x-4)(x-2-3i)(x-2+3i) 
  
Edwin