f(x) = x^5 - 2x^4 - 2x^3 + x^2 + 4x + 4 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 a 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

### 1 Answer | Add Yours

`f(x) = x^5 - 2x^4 - 2x^3 + x^2 + 4x + 4`

It has (x-2) as a factor.

Rearrange the original equation and factorize as:

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

`=x^4(x-2)-2x^2(x-2)-3x(x-2)-2(x-2)`

`=(x-2)(x^4-2x^2-3x-2)`

`=(x-2)(x^4-2x^3+2x^3-4x^2+2x^2-4x+x-2)`

`=(x-2){(x^3(x-2)+2x^2(x-2)+2x(x-2)+1(x-2)}`

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

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

`=(x-2)^2{(x+1)(x^2-x+1)+2x(x+1)}`

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

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

Thus, the roots of f(x) are, `2, 2, -1, (-1+-sqrt(1^2-4*1*1))/(2*1)`

i.e. `2, 2, -1, (-1+-sqrt(-3))/2`

Therefore, the given function has 2 as a double root, another one real root(-1), and two complex roots.

**Hence option E) is the correct answer.**

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