PLEASE HELP FAST!!!!

Question

Finding the zeros of a polynomial function.

f(x)=x^5+x^3-2x^2-12x-8

Answer 1

P = +or- {1,2,4,8}

Q = +or- {1}

so p/Q = P = +or- {1,2,4,8}

f(-1) = (-1) + (-1) -2 +12-8 = 0,

so x=-1 is a solution and (x+1) is a factor

-1 | 1 0 1 -2 -12 -8

-1 1 -2 4 8

___________________________

1 -1 2 -4 -8 0

the quotient is x^4 - x^3 + 2x^2 - 4x - 8 = Q(x)

Q(-1) = 1 - -1 + 2 + 4 - 8 = 0,

so again, x=-1 is a solution and (x+1) is a factor

-1 | 1 -1 2 -4 -8

-1 2 -4 8

_________________________

1 -2 4 -8 0

x^3 - 2x^2 + 4x-8 = 0

x^2(x-2) + 4(x-2) = 0

(x^2+4)(x-2) = 0

x-2 = 0 or x^2 + 4 = 0

x=2 or x = +or-2i

check: f(2) = 32 + 8 - 8- 24 - 8 = 0

f(2i) = (2i)^5 + (2i)^3 - 2(2i)^2 - 12(2i) - 8 =

32i -8i +8 - 24i - 9 = 0

So x=2 and x=2i are solutions, which forces x=-2i

must also be a solution

the solution set is x = {-1,-1,2, 2i, -2i}