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.
If a polynomial equation has only real coefficients, then non-real 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 "non-real", 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 $\displaystyle x^3- 6x^2+ 11x- 6= (x- 1)(x- 2)(x- 3)= 0$ has three "complex" roots- they are 1, 2, and 3.
The equation $\displaystyle x^3- 4x^2+ 5x-2= (x- 1)^2(x- 2)= 0$ has three "complex" roots- they are 1 (of multiplicity 2) and 2.
Hello, anthonye!
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) x-intercepts.
Code:* * * * * * * * * * * * *
It can have one x-intercept.
(One real zero, two complex zeros.)
Code:| | * | | | * * * * *| * * | * * * | * | | ---o--+---------------- | | | * | |
It can have two x-intercepts.
(Two real zeros, one is repeated.)
Code:| | * | *| * * | * * | * * | * * ----o-----+------o--- | | * | |
It can have three x-intercepts.
(Three real zeros.)
Code:| | * | | |* * * | * ---o--+----o-------o---- | * * * | * | | * | |
But it cannot have no x-intercepts (three complex zeros).
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.
I think finally I understand your confusion.
Every real number is a complex number.
Every complex number has the form $\displaystyle a+bi$ where $\displaystyle a~\&~b$ are real numbers and $\displaystyle i^2=-1$.
If $\displaystyle b=0$ we say that $\displaystyle a+bi$ is a real number.
So in your question in the quote $\displaystyle 3=3+0i$ is said to be real.
In the polynomial $\displaystyle \sum\limits_{k = 0}^n {{\alpha _k}{z^k}}$ if $\displaystyle \forall k$ we have $\displaystyle \text{Im}(\alpha_k)=0$ [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.
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.