f(x) = x^5 - 6x^4 + 11x^3 - 2x^2 - 12x + 8 has (x -2) as a factor. Determine which ONE of the following is true about the set of roots for f(x).

a) 2 is a single root with 4 more Real and 0 Complex

b) 2 is a single root with 2 more Real and 2 Complex

c) 2 is single root with no Real and 4 Complex

d) 2 is a double root with 3 more Real and 0 Complex

e) 2 is a double root with 1 more Real and 2 Complex

f) 2 is a triple root with no Real and 2 Complex

g) 2 is a triple root with 2 more Real and 0 Complex

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`f(x)= x^5 - 6x^4 + 11x^3 - 2x^2 - 12x + 8`

Since (x-2) is a factor of f(x), then x=2 is one of the roots of this function. To determine its multiplicity, let's divide the polynomial by this root using synthetic division.

`2` `|` `1` `-6` `11` `-2` `-12` `8`

`2` `-8` `6` `8` `-8`

_____________________________

`1` `-4` `3` `4` `-4` `0`

Since the last number is 0, this proves that 2 is a root of f(x).

Then, divide the quotient by x=2 again.

`2` `|` `1` `-4` `3` `4` `-4`

`2` `-4` `-2` `4`

__________________________

`1` `-2` `-1` `2` `0`

Here, the last number is 0. Hence, x=2 is a root of f(x) again.

Then, divide the quotient by x=2 again.

`2` `|` `1` `-2` `-1` `2`

`2` `0` `-2`

_____________________

`1` `0` `-1` `0`

The last number is 0 too. So, x=2 is a root of f(x) again.

Also, divide the quotient by the same value of x.

`2` `|` `1` `0` `-1`

`2` `4`

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`1` `2` `3`

Notice that the last number is not zero. This indicates that 2 is no longer a root of the function. So, we do not have to divide the quotient by x=2 again.

Since x=2 is a root of f(x) three times this indicates that the factor of the function is:

`f(x)= x^5 - 6x^4 + 11x^3 - 2x^2 - 12x + 8`

`f(x)= (x -2)^3(x^2+0x -1)`

`f(x)=(x-2)^2(x^2-1)`

To determine the other roots, factor x^2-1.

`f(x)=(x-2)^3(x-1)(x+1)`

Base on these, the roots of f(x) are `x=2` with multiplicity of 3,

`x=1` , and `x=-1` .

**Hence, among the given choices, it is statement (g) that is true.**

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