Question 765222
<pre>
Complex roots are like 2+3i, -4+7i, -5-7i, 6-3i, i, 3i, etc.
They are roots with i's.  Real root have no i terms.

Since x<sup>5</sup>-6x<sup>4</sup>+11x<sup>3</sup>-2x<sup>2</sup>-12x+8 has (x-2) as a factor

2|1 -6 11 -2 -12  8
 |<u>   2 -8  6   8 -8</u>
  1 -4  3  4  -4  0

So we have factored f(x) as

f(x) = (x-2)(x<sup>4</sup>-4x<sup>3</sup>+3x<sup>2</sup>+4x-4) = 0

Since 2 is a factor of the last term (in absolute value),
it may be that 2 is a root more than once, so we try 2 again
on x<sup>4</sup>-4x<sup>3</sup>+3x<sup>2</sup>+4x-4

2|1 -4  3  4 -4
 |<u>   2 -4 -2  4</u>
  1 -2 -1  2  0

So we have factored f(x) as

f(x) = (x-2)(x-2)(x<sup>3</sup>-2x<sup>2</sup>-x+2) = 0

We can factor x<sup>3</sup>-2x<sup>2</sup>-x+2 by grouping:
             x<sup>2</sup>(x-2)-1(x-2)
             (x-2)(x<sup>2</sup>-1)
             (x-2)(x-1)(x+1)

So the final factorization is

f(x) = (x-2)(x-2)(x-2)(x-1)(x+1)

f(x) = (x-2)<sup>3</sup>(x-1)(x+1)

So 2 is a root of multiplicity 3 (a triple root, 
1 is a single root, and -1 is a single root.

So g is the correct answer.

Edwin</pre>