Zeroes of a polynomial.?

Question

I don't know how to find the zeroes of complex polynomials like this:

x^4 - 2x^3 - 2x^2 - 2x - 3

Could you help show me how? Thanks.

Answer 1

Rational Zero theorem
Take the factors of the last( non x) / factors of leading coeffienct

In this case Factor of 3 - or + 1, - or + 3 / - or + 1
So the zero can be -3,-1,1,3

Graph and see which is right -1 and 3 looks right so uses synthetic

-1 1 -2 -2 -2 -3
-1 3 -1 3
1 -3 1 -3 0 (with =0 at end if it is a factor)

x^3-3x^2+x-3 (now solve 3)

3 1 -3 1 -3
3 0 3
1 0 1 0

Now your eq is x^2+1
x^2+1=0
x^2=-1
x=i, -i (with are not real numbers but imaginary)

Answer 2

first, try to get x=1 into the poly. if equals to zero so x=1 is a zero of the poly. but it is not.
try x=-1 and it is. now divide the original poly with (x+1):
(x^4 - 2x^3 - 2x^2 - 2x - 3):(x+1)= x^3-3x^2+x-3
now, use the "trinom" and make the answer to (x^2+1)(x-3) and now the original poly can written as:
(x+1)(x-3)(x^2+1)
the values of x that make the poly equals to 0 are: -1, 3
Amir

Answer 3

if x=-1
then
1+2-2+2-3=0
hence it is a root

use Horner method to find the other roots
x^3 -3x^2 +x - 3=0

if x=3 then
27-27+3-3=0
Hence it is a root

x^2+3=0

no other real roots.
Answer: -1 and 3

Hint: use trial and error method and Horner Rule

Answer 4

f(-1)= (-1)^4-2(-1)^3-2(-1)^2-2(-1)-3
=1+2-2+2-3=0
therefore by remainder theorem (x+1) is a root
Factorizing
x^4+x^3-3x^3-3x^2+x^2+x-3x-3=0
x^3(x+1)-3x^2(x+1)+x(x+1)-3(x+1)=0
(x+1)(x^3-3x^2+x-3)=0
One zero is at x=-1. There are 3 other zeros.

Answer 5

my freind if you are see the last digit number -3 find all factors of -3 and test them like 1,-1,3,-3 if any of those are correct then you zero is that goood luck if you still can find it PM me Ok

Answer 6

use Matlab, roots([1 -2 -2 -2 -3]).
Or any other math tool.