Question

HELP!!
Identify all the zeros of the polynomial function f(x)= x^3-4x^2+6x-4

Answer 1

for polynomials of order 3, we usually start evaluating the polynomial (call it p) in the following values: 1, -1, 2, -2, 3, -3.
if you find a value x in those for which p(x)=0 it's cool, you use euclidian division (or horner's technique) to factorize.

Answer 2

Ok

Answer 3

I got -1, 2i, and -2i

Answer 4

what is the root you found by testing ±1, ±2,..? I found 2.

Answer 5

I had some help me with it, but thats what they got, is it right?

Answer 6

hmm. 2 is root: p(2) = 2^3 - 4 * 2^2 + 6 * 2 - 4 = 8 - 16 + 12 - 4 = 0.
Now: euclidian division (divide p(x) by (x-2))

Answer 7

\(x^3 - 4 x^2 + 6x - 4 = (x-2) (x^2 - 2x + 2))\).
The other roots of p(x) are the roots of \(x^2-2x+2\). Maybe they are 2i and -2i, i did not check.

Answer 8

Ok what did you get ?

Answer 9

i compute as i write.. so, \(\Delta=2^2-4\cdot2\cdot 1 = -4 = (2i)^2\).
Ok the roots are \((2\pm 2i)/2 = 1\pm i\).

Answer 10

ok

Answer 11

if the question is "find the real roots" then you don't need the two other roots of course..
I think my anwer is correct (checked mentally .. :| )

Answer 12

So 2, 1 + i, 1 - i?

Answer 13

yes

Answer 14

or 1, 1 + 2i, 1-2i

Answer 15

1+i is ok since (1+i)^2 = 2i, so: p(1+i) = 2i - 2(1+i) + 2 = 2i-2-2i+2 = 0.
therefore 1-i is ok too. ;)

Answer 16

so my 1 or 2 answer

Answer 17

it's: So 2, 1 + i, 1 - i.