You

**can** do synthetic division by complex roots. It's a little tricky, but it can be done.

First, divide $\displaystyle x^4 - 9x^3 + 21x^2 +21x -130$ by

**3 - 2i**. We know that the remainder will be zero:

Code:

3 - 2i| 1 -9 21 21 -130
------- 3-2i -22+6i 9+20i 130
--------------------------------
1 -6-2i -1+6i 30+20i 0

Now, let's divide the quotient by

**3 + 2i**, the conjugate of 3 - 2i:

Code:

3 + 2i| 1 -6-2i -1+6i 30+20i
------- 3+2i -9-6i -30-20i
---------------------------
1 -3 -10 0

So you're left with the quadratic

$\displaystyle x^2 - 3x - 10 = 0$

which factors into

$\displaystyle (x + 2)(x - 5) = 0$

The two remaining roots are -2 and 5.

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