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If you are asked to write a polynomial function of least degree with real coefficients and with zeros of 2 and i square roots of seven what would be the degree of the polynomial also wright equation?
3y2-5xyz yay i figured it out!!!!
Is it true that a polynomial's real roots are the values at which the graph of a polyomial meets the x-axis?
Yes.
How is the Degree of a polynomial function related to range?
If the domain is infinite, any polynomial of odd degree has infinite range whereas a polynomial of even degree has a semi-infinite range. Semi-infinite means that either the range has a real minimum but no maximum (ie maximum = +infinity) or that it has a real maximum but no minimum (ie minimum = -infinity).
How do you determine the number of real roots in a polynomial?
For a general polynominal, the cubic, quartic, and greater formulæ are too hellishly hard to work with, so you would need to plot the function or use Newton's/somesuch method to count the real roots by hand. If the polynomial has integral roots, you can use synthetic division to peel off the degrees to see if they factor wholely into binominals; then all roots will be real and explicit. Good luck:
How many roots can a quadratic function have?
The answer is two. Despite its name seems to suggest something to do with four, in a quadratic equation the unknown appears at most to the power of two and so is said to be of second degree. The theorem than pertains here is that the number of roots an equation has is equal to its degrees. However, some of the roots can be repeated - an nth degree equation need not have n different roots. Also the roots do not have to be real. However complex roots ( no real) come in pairs so an equation of odd degree must have at least one real root. A quadratic possibly has no real roots.
Is it true that the degree of polynomial function determine the number of real roots?
Sort of... but not entirely. Assuming the polynomial's coefficients are real, the polynomial either has as many real roots as its degree, or an even number less. Thus, a polynomial of degree 4 can have 4, 2, or 0 real roots; while a polynomial of degree 5 has either 5, 3, or 1 real roots. So, polynomial of odd degree (with real coefficients) will always have at least one real root. For a polynomial of even degree, this is not guaranteed. (In case you are interested about the reason for the rule stated above: this is related to the fact that any complex roots in such a polynomial occur in conjugate pairs; for example: if 5 + 2i is a root, then 5 - 2i is also a root.)
How many real roots can a fourth degree polynomial have?
Upto 4. If the coefficients are all real, then it can have only 0, 2 or 4 real roots.
What is the relationship between the degree of a polynomial and the number of roots it has?
In answering this question it is important that the roots are counted along with their multiplicity. Thus a double root is counted as two roots, and so on. The degree of a polynomial is exactly the same as the number of roots that it has in the complex field. If the polynomial has real coefficients, then a polynomial with an odd degree has an odd number of roots up to the degree, while a polynomial of even degree has an even number of roots up to the degree. The difference between the degree and the number of roots is the number of complex roots which come as complex conjugate pairs.
How many real roots do we have if the polynomial equation is in degree six?
Such an equation has a total of six roots; the number of real roots must needs be even. Thus, depending on the specific equation, the number of real roots may be zero, two, four, or six.
How can you quickly determine the number of roots a polynomial will have by looking at the equation?
In the complex field, a polynomial of degree n (the highest power of the variable) has n roots. Some of these roots may be multiple roots. However, if the domain is the real numbers (or a subset) then there is no easy way. The degree only gives the maximum number of roots - there may be no real root. For example x2 + 1 = 0.
If you are asked to write a polynomial function of least degree with real coefficients and with zeros of 2 and i square roots of seven what would be the degree of the polynomial also wright equation?
3y2-5xyz yay i figured it out!!!!
The polynomial given below has rootss?
You forgot to copy the polynomial. However, the Fundamental Theorem of Algebra states that every polynomial has at least one root, if complex roots are allowed. If a polynomial has only real coefficients, and it it of odd degree, it will also have at least one real solution.
How to tell if there are no real roots?
The real roots of what, exactly? If you mean a square trinomial, then: If the discriminant is positive, the polynomial has two real roots. If the discriminant is zero, the polynomial has one (double) real root. If the discriminant is negative, the polynomial has two complex roots (and of course no real roots). The discriminant is the term under the square root in the quadratic equation, in other words, b2 - 4ac.
What is the greatest number of real roots a polynomial of degree 2 can have?
A real root is when a quadratic equation, or the graph of a polynomial, crosses the x axis, or when the y coordinate is equal to 0. On any polynomial to the degree of two, when graphed the line follows a smooth arc in the shape of a "U" or and upside down "U". Since there are only two prongs to the parabola, or arc, it can only cross the x axis twice, if at all. So there can only be 2 real roots.
How can a rational equation have more than one solution?
A rational equation can be multiplied by the least common multiple of its denominators to make it into a polynomial equation. The degree of this polynomial is the highest power (of the variable) that appears in it. It can be proven that a polynomial of degree n must have n roots in the complex domain. However, there may be fewer roots in the real domain. This is because if the coefficients are real then there may be pairs of complex roots [conjugates] which will not count as real roots. Also, there may be identical roots of multiple order. For example, x4 - 1 = 0 has 4 complex roots. These are 1, -1, i and -i where i is the imaginary root of -1. There are only 2 real roots -1 and +1. x4 = 0 has 4 multiple roots, each of which is 0. Thus x = 0 is a root of multiplicity 4.
Are skew symmetric roots purely real or purely imaginary?
They can be either. If they are roots of a real polynomial then purely imaginary would be symmetric and only real roots can be skew symmetric.
How do you find out the number of imaginary zeros in a polynomial?
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).
