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Descartes' rule of signs (see related link) can help you determine the maximum number of real roots. If the polynomial is odd powered, then there will be at least one real root. Any even powered polynomial can be factored into a bunch of quadratics [though they may not be rational or even pretty], and any odd-powered polynomial can be factored into a bunch of quadratics and one linear (this one would have the real root). So the quadratics may have pairs of real or complex roots (having an imaginary component).
To clarify, when I say complex, I'm referring to the fact that there will be an imaginary component to the root, because actually the real numbers is a subset of the set of complex numbers.
The order of the polynomial will tell you how many roots it will have. If you can graph the polynomial, then you can see if it crosses the x axis. If it is a 5th order polynomial, and crosses the x axis 3 times, then there are 3 real roots (the other two roots are complex).
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Is 13 a polynomial If it is find its degree and classify it by the number of its terms?
13 is not a polynomial.
To find the factors of a polynomial from its graph follow this rule If the number is a root of a polynomial then x - b is a factor?
a
Why do you say that pi is an imaginary number?
If I ask Answers™ "what is pi squared?" I find "It is approximately equal to 3.14 but in reality pi is an imaginary number that has no end." The answer also goes on to tell me that imaginary numbers cannot be multiplied by themselves. Now i must see what y'all have to say about imaginary numbers...
How do you find the greatest common factor of a polynomial?
A single polynomial cannot have a greatest commonfactor. There is nothing that it will be in common with!
How do you find cube roots of imaginary numbers?
For a pure imaginary number: {i = sqrt(-1)} times a real coefficient {r}, you have i*r. The cube_root(i*r) = cube_root(i)*cube_root(r), so find the cube root of r in the normal way, then we just need to find the cube root of i. For any cubic function (which has a polynomial, in which the highest term is x3) will always have 3 roots. There are 3 values, which when cubed will equal the imaginary number i:-i will do it: (-i)3 = (-i)2 * (-i) = -1 * (-i) = iThe other two are complex: sqrt(3)/2 + i/2 and -sqrt(3)/2 + i/2If you cube either of the two complex binomials by multiplying out, you will end up with 0 + i as the answer in both cases.Note: the possible roots for any cubic are: 3 real roots, or 1 real root and 2 complex root, or 1 pure imaginary root, and 2 complex roots.For your original question, if you want to stay in the pure imaginary domain, then you can use: Cube_root(i*r) = -i * cube_root(r) to find an answer.
