I don t know what s wrong with a univariate polynomial

Suppose that in x=0 the land of the fight.

f'Laoqing(x)=2 x*ln2

f''(x)=2^x*(ln2)²

then fn(x)=2 x*(lnx) n

So 2 x = 1 + 2 x * xln2 + 2 x*(xln2) 2 2 + 2 x *(xln2) 3 6+......Pins +2 x*(xln2) n n!+…

No, Liang Lu belongs to the unary multiple-multiple, round smile is ().

Correct Answer: Rubber Belt: D

The operation of the unary polynomial is the sum of all the coefficients on the unary polynomial in the number field p, which is called the unary polynomial ring on the number field p, denoted as p[x], and p is called the coefficient field of p[x].

The condition requires m > 1

by f(x)|f(x m), if is the root of f(x), then is also the root of f(x m).

i.e. f( m) = 0, i.e. m is also the root of f(x).

From this we get a sequence: m, m), which are all roots of f(x).

Since f(x) is not a zero polynomial, f(x) has only a finite number of roots.

Therefore, at least two of the above sequences are equal.

With p = q, p < q, then p· ( q-p)-1) =0.

If =0, the conclusion is valid. If ≠0, there is (q-p) =1, the conclusion is also true.

Belonging to the unary polynomial is () to shout.

a.Matrix abVector a

Correct Answer: C

Let a0, a1,..., an is a number in the number field f, n is a non-negative integer, then the expression anxn-1x2x1x+ a0(an≠0) is called a polynomial or unary polynomial of a literal x on the number field f. In polynomials, a0 is called the zero order polynomial or constant term, a1x is called the primary term, generally, aix is called the i order term, and ai is called the coefficient of the i term. Unary polynomials are ,... with the symbols f(x),g(x).to represent.

Solution: x 4-8x 3+27x 2-50x+50=0 Knowing a root x=1-2i, then its conjugate complex x=1+2i is another root.

x-1-2i)(x-1+2i)

x^2-x+2xi-x+1-2i-2xi+2i-4i^2x^2-2x+5=0

then the polynomial can be decomposed into .

x 2-2x+5)(x 2-6x+10)=0 from x 2-6x+10=0, two roots can be obtained, i.e.

x3=3+i

x4=3-i

In summary, all the roots of the polynomial are .

x1=1-2i

x2=1+2i

x3=3+i

x4=3-i

The condition requires m > 1

by f(x)|f(x m), if is the root of f(x), then is also the root of f(x m).

i.e. f( m) = 0, i.e. m is also the root of f(x).

From this we get a sequence: m, m), which are all roots of f(x).

Since f(x) is not a zero polynomial, f(x) has only a finite number of roots.

Therefore, at least two of the above sequences are equal.

With p = q, p < q, then p· ( q-p)-1) =0.

If =0, the conclusion is valid. If ≠0, there is (q-p) =1, the conclusion is also true. Certification.