Hi;
If I have a 3rd degree polynomial it has three roots(zeros) of which three can be complex.
Why a complex number has two parts real and imaginary and I thought these came in conjugate pairs?
Thanks.
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Hi;
If I have a 3rd degree polynomial it has three roots(zeros) of which three can be complex.
Why a complex number has two parts real and imaginary and I thought these came in conjugate pairs?
Thanks.
What still confised.
If a polynomial equation has only real coefficients, then nonreal roots must come in conjugate pairs. If you had a third degree equation with real coefficients, it must have either three or one real root.
However notice that I said "nonreal", not "complex". The set of all complex number includes the real numbers. That is why we can say that any nth degree polynomial equation has n (counting multiplicity) complex roots. The equation has three "complex" roots they are 1, 2, and 3.
The equation has three "complex" roots they are 1 (of multiplicity 2) and 2.
So a true complex number would be a  bi or a + bi.
But a complex number can also be say 1,2,3...ect it's just that the imaginary part is zero?
I could also have bi as a complex number where the real part is zero?
Hello, anthonye!
Quote:
If I have a 3rd degree polynomial, it has three roots (zeros) of which three can be complex.
. . This is impossible.
Consider the graph of a cubic function.
The zeros of the polynomial are the (real) xintercepts.
Code:*
* *
* *
* * *
* *
* *
*
It can have one xintercept.
(One real zero, two complex zeros.)
Code:
 *


 * *
* *
* * *
 * *
*  *


o+



* 

It can have two xintercepts.
(Two real zeros, one is repeated.)
Code:
 *

* *
*  *
*  * *
 * *
o+o


* 

It can have three xintercepts.
(Three real zeros.)
Code:
 *


* *
*  *
o+oo
 * *
*  *


* 

But it cannot have no xintercepts (three complex zeros).
Ok thanks for taking the time Soroban I'll look later.
Yes, to each of those. "bi as a complex number where the real part is zero" would be a pure imaginary number. We can say that "every complex number is the sum of a real number and a pure imaginary number" but we have to count 0 as both real and complex. That's not such an unreasonable thing to do. In the "complex plane", the real and imaginary axes intersect at 0: 0+ 0i.
Ok getting there now.
This is what I have:
For the equation x^4  3x^3  x^2  x + 3 = 0
Find the number of complex roots,The possible number of real roots
and the possible number of rational roots.
complex roots = 4 (The reason for my answer is complex numbers are almost any numbers)
real roots = 0,2,4 ( the reason for my answers are 0 the answer may all be complex) and this is where i'm confused
real numbers are on the number line 3,2,1,0,1,2,3ect But if a complex number is say 3(imaginary part=0) is 3 complex or
real?
Thanks.
Yes I get that thats a imaginary number but it must also be complex with the real part being zero yes/no?
I think finally I understand your confusion.
Every real number is a complex number.
Every complex number has the form where are real numbers and .
If we say that is a real number.
So in your question in the quote is said to be real.
In the polynomial if we have [the imaginary part is zero] we say that polynomial has real coefficients. Therefore, any complex roots (meaning the imaginary part is not zero) must occur in conjugate pairs.
Ok thanks again all have to go now back later(I'm starvin).
So going back to the equation x^4  3x^3  x^2  x + 3 = 0.
How can I tell from just the equation how many roots of each type it will have?
I know I can use the rational root theorem and synthetic division to find the first two roots
and then the quadratic formula to find the last two that will tell me what they are and how many of each type
but can I tell what type and how many before I know what their values will be?
Thanks.