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How many terms does the polynomial have?
As many as you like. A polynomial in 1 variable, and of degree n, can have n+1 terms where n is any positive integer.
Is second-degree polynomial is a quadratic polynomial?
Yes, any second-degree polynomial is quadratic. Degree 0 - constant (8) Degree 1 - linear (n) Degree 2 - quadratic (n^2) Degree 3 - cubic (n^3) Degree 4 - fourth degree (n^4) Degree 5 - fifth degree (n^5) Degree 6 - sixth degree (n^6) and so on............ Also a degree I find funny is the special name for one hundredth degree. Degree 100 - hectic (n^100)
How do you rewrite an polynomial and standard form?
A polynomial, of degree n, in standard form is:anxn + an-1xn-1 + ... + a1x+ a0 = 0 where n is an integer and the ai are constants.The answer about how to rewrite a polynomial depends on the form that it is given in.A polynomial, of degree n, in standard form is:anxn + an-1xn-1 + ... + a1x+ a0 = 0 where n is an integer and the ai are constants.The answer about how to rewrite a polynomial depends on the form that it is given in.A polynomial, of degree n, in standard form is:anxn + an-1xn-1 + ... + a1x+ a0 = 0 where n is an integer and the ai are constants.The answer about how to rewrite a polynomial depends on the form that it is given in.A polynomial, of degree n, in standard form is:anxn + an-1xn-1 + ... + a1x+ a0 = 0 where n is an integer and the ai are constants.The answer about how to rewrite a polynomial depends on the form that it is given in.
How do you polynomial whose zeros are given?
If a polynomial p(x), has zeros at z1, z2, z3, ... then p(x) is a multiple of (x - z1)*(x - z2)*(x - z3)... To get the exact form of p(x) you also need to know the order of each root. If zk has order n then the relevant factor in p(x) is (x - zk)n
Why does 15 not belong in the number series of 15 2 8 13 16?
Of course it does.Given any set of n numbers, it is ALWAYS possible to find a polynomial of degree (n-1) such that the polynomial generates those numbers.In this case, tryUn = (19n4 - 270n3 + 1373n2 - 2826n + 2064)/24 for n = 1, 2, 3, 4 etc.So, the question is misguided.Of course it does.Given any set of n numbers, it is ALWAYS possible to find a polynomial of degree (n-1) such that the polynomial generates those numbers.In this case, tryUn = (19n4 - 270n3 + 1373n2 - 2826n + 2064)/24 for n = 1, 2, 3, 4 etc.So, the question is misguided.Of course it does.Given any set of n numbers, it is ALWAYS possible to find a polynomial of degree (n-1) such that the polynomial generates those numbers.In this case, tryUn = (19n4 - 270n3 + 1373n2 - 2826n + 2064)/24 for n = 1, 2, 3, 4 etc.So, the question is misguided.Of course it does.Given any set of n numbers, it is ALWAYS possible to find a polynomial of degree (n-1) such that the polynomial generates those numbers.In this case, tryUn = (19n4 - 270n3 + 1373n2 - 2826n + 2064)/24 for n = 1, 2, 3, 4 etc.So, the question is misguided.
How many terms can a polynomial have?
As many as you like. The highest power of the variable in question (usually x) defines the degree of the polynomial. If the degree is n, the polynomial can have n+1 terms. (If there are more then the polynomial can be reduced.) But there is NO LIMIT to the value of n.
How many terms does the polynomial have?
As many as you like. A polynomial in 1 variable, and of degree n, can have n+1 terms where n is any positive integer.
How do you solve this For a polynomial of degree n express the coefficient of x to the n-1 power and the constant coefficient in terms of the zeros of the polynomial?
If the roots are r1, r2, r3, ... rn, then coeff of x^(n-1) = -(r1+r2+r3+...+rn) and constant coeff = (-1)^n*r1*r2*r3*...*rn.
Is second-degree polynomial is a quadratic polynomial?
Yes, any second-degree polynomial is quadratic. Degree 0 - constant (8) Degree 1 - linear (n) Degree 2 - quadratic (n^2) Degree 3 - cubic (n^3) Degree 4 - fourth degree (n^4) Degree 5 - fifth degree (n^5) Degree 6 - sixth degree (n^6) and so on............ Also a degree I find funny is the special name for one hundredth degree. Degree 100 - hectic (n^100)
How do you rewrite an polynomial and standard form?
A polynomial, of degree n, in standard form is:anxn + an-1xn-1 + ... + a1x+ a0 = 0 where n is an integer and the ai are constants.The answer about how to rewrite a polynomial depends on the form that it is given in.A polynomial, of degree n, in standard form is:anxn + an-1xn-1 + ... + a1x+ a0 = 0 where n is an integer and the ai are constants.The answer about how to rewrite a polynomial depends on the form that it is given in.A polynomial, of degree n, in standard form is:anxn + an-1xn-1 + ... + a1x+ a0 = 0 where n is an integer and the ai are constants.The answer about how to rewrite a polynomial depends on the form that it is given in.A polynomial, of degree n, in standard form is:anxn + an-1xn-1 + ... + a1x+ a0 = 0 where n is an integer and the ai are constants.The answer about how to rewrite a polynomial depends on the form that it is given in.
Can there be two patterns in a sequence?
Yes, there can be infinitely many. Given a sequence of n numbers, it is always possible to fit a polynomial of degree (n-1) to it. That polynomial is one posible pattern.Then suppose the sequence is extended by adding an (n+1)thnumber = k. You now have a sequence of n+1 numbers and there is a polynomial of degree n that will fit it. For each of an infinite number of values of k, there will be a different polynomial of degree n. Next add another number, l. There will now be an infinite number of polynomials of degree n+1. And this process can continue without end.And these are only polynomial functions. You can have other rules - for example, sums of sines and cosines (see Fourier transformations if you are really keen and able).
What is the standard form for a polynomial?
The standard form of a polynomial of degree n is anxn + an-1xn-1 + ... + a1x + a0 where the ai are constants.
Is there any connection between Evariste Galois's group theory and symmetry?
There sure is, and a major connection at that.Consider a finite set of n elements. The symmetric group of this set is said to have a degree of n. The symmetric group of degree n (Sn) is the Galois group of the general polynomial of degree n. In order for there to be a formula involving radicals that solve the general polynomial of degree n, such as the quadratic equation when n = 2, that polynomial's corresponding Galois group must be solvable. S5 is not a solvable group. Therefore, the Galois group of the general polynomial of degree 5 is not solvable. Thus the general polynomial of degree 5 has no general formula to solve it using radicals.This was huge result, and one of the first real applications, for group theory, since that problem had stumped mathematicians for centuries.
What is schur cohn stability?
A test is given for deciding whether a prescribed real polynomial of degree n has all its zeros inside the unit circle. The test involves examination of the sign of various linear combinations of the polynomial coefficients, as well as examination to check positive definiteness of a symmetric matrix of dimensions +n X f n when n is even, and f ( n + 1) X +(n 4- 1) or f(n - 1) X f(n. - l j
How do you find the roots of a polynomial of graphed points?
Join the points using a smooth curve. If you have n points choose a polynomial of degree at most (n-1). You will always be able to find polynomials of degree n or higher that will fit but disregard them. The roots are the points at which the graph intersects the x-axis.
How do you polynomial whose zeros are given?
If a polynomial p(x), has zeros at z1, z2, z3, ... then p(x) is a multiple of (x - z1)*(x - z2)*(x - z3)... To get the exact form of p(x) you also need to know the order of each root. If zk has order n then the relevant factor in p(x) is (x - zk)n
Why does 15 not belong in the number series of 15 2 8 13 16?
Of course it does.Given any set of n numbers, it is ALWAYS possible to find a polynomial of degree (n-1) such that the polynomial generates those numbers.In this case, tryUn = (19n4 - 270n3 + 1373n2 - 2826n + 2064)/24 for n = 1, 2, 3, 4 etc.So, the question is misguided.Of course it does.Given any set of n numbers, it is ALWAYS possible to find a polynomial of degree (n-1) such that the polynomial generates those numbers.In this case, tryUn = (19n4 - 270n3 + 1373n2 - 2826n + 2064)/24 for n = 1, 2, 3, 4 etc.So, the question is misguided.Of course it does.Given any set of n numbers, it is ALWAYS possible to find a polynomial of degree (n-1) such that the polynomial generates those numbers.In this case, tryUn = (19n4 - 270n3 + 1373n2 - 2826n + 2064)/24 for n = 1, 2, 3, 4 etc.So, the question is misguided.Of course it does.Given any set of n numbers, it is ALWAYS possible to find a polynomial of degree (n-1) such that the polynomial generates those numbers.In this case, tryUn = (19n4 - 270n3 + 1373n2 - 2826n + 2064)/24 for n = 1, 2, 3, 4 etc.So, the question is misguided.
