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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.
