Question 279010
<pre><font size = 4 color = "indigo"><b>
First write {{{P(x)=x^3 - 11x + 20}}} as 

{{{P(x)=x^3 +0x^2- 11x + 20}}}

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

{{{(2-i)(2-i)=4-2i-2i+i^2=4-4i+i^2=4-4i+(-1)=3-4i}}}

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

{{{(-8-4i)(2-i)=-16+8i-8i+4i^2=-16+4i^2=-16+4(-1)=-16-4=-20}}}

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 

{{{P(x)=x^3 - 11x + 20}}}

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 

{{{P(x)=x^3 - 11x + 20}}}

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</pre>