SOLUTION: Find all other zeros of {{{P(x)= x^3 - 11x + 20}}} , given that 2-i is a zero. (If there is more than one zero, separate them with commas.)

First write  as 



Then you can use synthetic division with the zero 2-i

2-i | 1    0     -11     20
    !      2-i
     ----------------------
      1    2-i

Now we have to stop and multiply 2-i by 2-i



Continuing with the synthetic division:

2-i | 1    0     -11     20
    !      2-i     3-4i
     ----------------------
      1    2-i    -8-4i

Now we have to stop again and multiply -8-4i by 2-i



Continuing with the synthetic division:

2-i | 1    0     -11     20
    |      2-i     3-4i -20
     ----------------------
      1    2-i    -8-4i   0

So we have factored 



as

P(x) = [x - (2-i)][x² + (2-i)x + (-8-4i)]

Now since 2-i is a zero, so is its conjugate is 2+i

So we use synthetic division with the second factor
and zero 2+i

2+i | 1    2-i   -8-4i
    |      2+i    8+4i     
     -----------------
      1    4       0

Now we have completed the factoring of 



as

P(x) = [x - (2-i)][x - (2+i)](x + 4)

So now we see that the third zero is -4.

So the three zeros of P(x) are 2-i, 2+i, and -4.

Edwin