Just for the heck of it.........?

Question

who ever can answer this right gets 10 pionts......

Given a function and one of its zeros, find all the zeros of the function.
f(x)=x^3 +2x^2 -3x+20; -4

go ahead give it a try......see if your smart....^_-

note: i don't need this for a class or anything, i just want to give people a challenge..

Answer 1

x+4 is a root
therefore
......._x^2-2x_+5___________
x+4/ x^3 +2x^2-3x+20
........x^3 +4x^2
.......------------------
...............-2x^2-3x
...............-2x^2-8x
............--------------------
......................5x+20
....................5x+20
...........----------------------
..................xxxxxxxxxx
0=(x+4) (x^2-2x+5)
x=[2+/-√(4-20)]/2
x=[2+/-4i]/2
x=1+/- 2i
Roots: -4,1+2i, 1-2i

Answer 2

f(x)=x³ +2x² -3x+20 : -4
f(x) = x³ + 2x² - 3x -5
f(0) = -5 (0,-5)
y = 0
x³ + 2x² - 3x -5 = 0
x(x² + 2x - 3 - 5) = 0
x² + 2x -8 = 0
x = 0
x' = 4
x" = -2
The answers: 0, 4, -2.
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Answer 3

-4 is the only real root. there's no sign change. the other 2 are imaginary numbers