Ok thank you again plato it's a question on a web site and it say that it has four complex roots
thats what was confusing me.
Look at this solution.
The polynomial has degree four and has all real coefficients. That tells us that if there are any complex roots (i.e. the imaginary parts non-zero) they must occur in conjugate pairs. Thus there could be zero, two, or four complex roots. In fact there are two. There are also two irrational roots. See the web link above.
Are you sure that you copied the equation correctly?
You can use WolframAlpha to solve such equations.
Hi;
Question state the number of complex roots,the possible number of real roots
of x^3 + 4x^2 + 5x -1 = 0
complex roots = 3 why(complex numbers include real numbers)
real roots = 1 or 3 why(1 in case two are complex,3 non may be complex)
are my statements correct?
Yes, that is true.
Well, not just integers, of course, numbers like " ", " ", etc. are real numbers. And you can say a little bit more. By "DeCarte's rule of signs", there can be either 0 or 2 positive real roots (because the sign changes at " " and at "+3"- two sign changes. DesCarte's rule of signs says the number of positive real roots is at most the number of sign changes but may be less by an even number. Now, replacing "x" by "-x" swaps positive and negative roots and which also has 0 or 2 sign changes.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
Both. The set of all real numbers is a subset of the set of all complex numbers.But if a complex number is say 3(imaginary part=0) is 3 complex or
real?
For the question about rational roots, you can use the "rational roots theorem": if the rational number , with m and n integers, of the polynomial equation , all coefficients being integers, then n must evenly divide the "leading coefficient", and m must evenly divide the "constant term", .Thanks.
Here, so the "leading coefficient" is 1 which is evenly divisible only by 1 and -1. The "constant term" is 3 which is evenly divisible only by 1, -1, 3, and -3. That is, the only possible rational roots are the integers 1, -1, 3, and -3. You can put each into the polynomial to see which, if any, actually satisfy the equation.
I think that you have misunderstood the whole point of this discussion.
Here is a direct quote: " the cubic polynomial has two complex roots"
Now that polynomial has three roots, one real and two complex.
As an accident of history, the language of roots of a polynomial we use the words complex root to mean a complex number with a nonzero imaginary part. Thus , has at least one real root and at most two complex roots.
NOTE: above we are talking about complex roots, not complex numbers as such.
Yes, because non-real solutions to an equation with real coefficients must come in complex conjugate pairs, there must be either 2 or no non-real solutions. That leaves either 1 or 3 real roots. And we can say if there only one real root it must be positive. If all three roots are real, one is positive and the other two negative.
Hi;
Just need my work checking(I'm sure it's all correct now)
For each equation state the number of complex roots,the possible number or real roots
and the possible rational roots.
1) -x^4 = 0 (complex roots 0 or2 or 4)
(real roots 2 or 4)
(rational roots 0)
2) 2x^5 - 4x^4 - 4x^2 + 5 = 0 (complex roots 0 or 2 or 4)
(real roots 1 or 3 or 5)
(rational roots + or - 5/2 or 1/2 or 1 or 5)
3)4x + 8 = 0 (complex roots 0)
(real roots 1)
(rational roots + or - 1/4 or 1/2 or 1 or 2 or or 8)
4)-2x^6 - x^2 + x - 7 = 0 (complex roots 0 or 2 or 4 or 6)
(real roots 2 or 4 or 6)
(rational roots + or - 1 or 1/2 or 7 or 7/2)
5)x^10 + x^8 - x^4 + 3x^2 - x + 1 = 0 (complex roots 0 or 2 or 4 or 6 or 8 or 10)
(real roots 0or 2 or 4 or 6 or 8 or 10)
(rational roots + or - 1)