polynomial什么意思

Which of the following CANNOT be a root of a polynomial in x of the form 9x^5+ax^3+b,where a and b are integers?
(a)-9 (b) -5 (c)1/4 (d)1/3 (e)9

How many real roots does the polynomial 2x^5+8x-7 have?
(A)None (B)One(C)Two(D)Three(E)Five

一道数学题（英文的!）
Given that P(x) is the least degree polynomial with rational coefficients such that P (√2+√3 ) =√2 ,
and then P(10)=

一道英语数学题
A polynomial of degree 3 that has 2,2 and 3 as roots can be written as
x3 + a x2 + b x + c where a,b and c are real numbers.Find a + b + c .a.-3
b.-2
c.-1
d.-6
e.0

Assume that p is a polynomial function on the set of real numbers.If p(0)=p(2)=3 and p'(0)=p'(2)=-1,then ∫(0,2)xp''(x)dx=?
(a)-3(b)-2(c)-1(d)1(e)2

一道数学英文题
Suppose a polynomial of degree 5 with rational coefficients has 3+根号3,1+2i,-1as roots.find all the roots of the polynomial.

a degree 3 polynomial that has zeros -1,1 and 6 and in which the coefficient of x^2 is -12.

Solve the polynomial equation:
-12x⁴- 8x³ + 49x² + 39x - 18 = 0

find a third degree polynomial with real coefficients that has 2,3-i as zeros

一道数学英文题
Find a polynomial of degree 4with real coefficients and i as a root of multiplicity 1 and-3 with multiplicity 2.

是一道英文的
For any polynomial fuction of degree n,the nth differences:
are equal(or constant)
have the same sign as the leading coefficient
are equal to a(nx(n-1)...x2x1),where a is the leading coefficient
希望大家能帮我解释下T T
我一点都看不懂
给到例题.
fourth difference=144
degree 4,leading coefficient 6

Every polynomial with complex coefficients of degree n has n zeros.Find all of the solutions to the equation:(x-2)(x-3)(x-4)(x-5) = 360

c++ 多项式 Polynomial
#include
using namespace std;
struct Polynomial {
int degree;
int *coeffs; // array of coefficients from
// lowest degree to highest degree
};
// reads the coefficients of a polynomial from standard input and
// creates a polynomial which it returns
Polynomial* readPoly();
// outputs a polynomial to standard output using the variable x
void outputPoly( Polynomial* p,char x );
// computes the sum of two polynomials
Polynomial* addPoly( Polynomial* a,Polynomial* b );
// computes the product of two polynomials
Polynomial* multPoly( Polynomial* a,Polynomial* b );
int main()
{
char x;
Polynomial *p;
readPoly();
outputPoly(p,x);
return 0;
}
Polynomial* readPoly()
{
int deg,*coefficient;
coutdeg;
coefficient = new int[deg+1]; // space for deg+1 coefficient
cout degree = deg;
p -> coeffs = coefficient;
}
//if the last degree is associated with a zero ,the output will show an extra "+" sign in the end
void outputPoly(Polynomial* p,char x)
{
x = 'x';
//char sign = '+';
for(int i= 0;idegree;i++){
string sign = "+";
if(p->coeffs[i] == 0)
{
continue;
}
if(i == 0)
{
if((p->coeffs[i+1]degree))
sign = "";
coutcoeffs[i]degree)) //if it is the last coefficent with number
sign = "";
cout

英文版的数学书看不懂
Carry out the following polynomial divisions.
题目大约就是一些不大好因式分解的多项式作为分子分母,从而组成的分式
没文化真是苦恼...

Solve the polynomial equation:
x^5 - 5x⁴ - 4x³+ 36x²+ 27x - 135 = 0

if p(x) denotes a polynomial of degree n such that P(k)=1/k for k=1,2,.,n+1,determine p(n+2).
怎么翻译 请大侠们帮忙

Write a formula for the third degree polynomial whose graph is given below
这道题是叫我把这个图像的函数表达式写出来吗?formula不是公式的意思吗?难道叫我写公式?

a degree 3 polynomial that has zeros -1,1 and 6 and in which the coefficient of x^2 is -12.

英语翻译
Conjecture 1.There exists a polynomial fk ∈ Q[x] with deg(fk) = ⌈(k−2)/2⌉ such that for a fixed nk (most probably nk = 2k) the equality lk(n) = fk(n) holds for n ≥ nk.
The aim of this paper is to confirm this conjecture by giving an explicit formula for the polynomial fk.For this purpose,we shall prove a new formula for the number of homomorphisms from Pn to Pk,which is the content of the following theorem.
Theorem 2.For any positive integers n and k,
|Hom(Pn,Pk)| = k × 2n−1
From the above theorem we are able to derive the following main result.
Theorem 3.If n ≥ 2k,then
Equivalently,the above formula can be rephrased as follows
When n ≥ 2k,Theorem 3 shows immediately that lk(n) is a polynomial in n of degree ⌈(k−2)/2⌉.This proves Conjecture 1.In particular,we have l1(n) = 1,l2(n) = 2,l3(n) = n,l4(n) = 2(n−1),l5(n) = 1 2n(n−1) and l6(n) = (n−1)(n−2),which coincide with the conjectured values in [5] after shifting the index by 1.
In the next section,we first recall some basic counting results about the lattice paths and then prove Theorem 2.In Section 3 we give the proof of Theorem 3.
2.The number of homomorphisms between paths
One can enumerate homomorphisms from Pn to Pk by picking a fixed point as image
of 1 and moving to vertices which are adjacent to this vertex,as
f ∈ Hom(Pn,Pk) ⇔ ∀x ∈ {1,...,n − 1} :{f(x),f(x + 1)} ∈ E(Pk).
Hence,one can describe all possible moves through the edge structure of the two paths.
For 1 ≤ j ≤ k,let
Obviously,we have
ON THE NUMBER OF CONGRUENCE CLASSES OF PATHS 3
Figure 1.A lattice path from (0,0) to (9,5) that stays between lines y = x and y = x − k + 1,where n = 15 and k = 11.

英语翻译
Explain whether the graph might represent a polynomial function that has zeros of order 2 or of order 3 ,就是这个问题,

C++问题:定义一个多项式类Polynomial,其实例为多项式：a0+a1x+a2x2+...+anxn
定义一个多项式类Polynomial,其实例为多项式：a0+a1x+a2x2+...+anxn,该类具有如下的接口：
class Polynomial
{ .
public:
Polynomial();
Polynomial(double coefs[],int exps[],int size);
//系数数组、指数数组和项数
Polynomial(const Polynomial&);
Polynomial();
Polynomial& operator=(const Polynomial&);
int degree() const; //最高幂指数
double evaluate(double x) const; //计算多项式的值
bool operator==(const Polynomial&) const;
bool operator!=(const Polynomial&) const;
Polynomial operator+(const Polynomial&) const;
Polynomial operator-(const Polynomial&) const;
Polynomial operator*(const Polynomial&) const;
Polynomial& operator+=(const Polynomial&);
Polynomial& operator-=(const Polynomial&);
Polynomial& operator*=(const Polynomial&);
};

When the polynomial F(x) of degree 3 is divided by x-2,the remainder is 1.
When F(x) is divided by x+3,the remainder is 6.
If F(x) is divided by (x-2)(x+3),the remainder is ax+b,where a and b care constants.
Find the values of a and b.
如何计算?请列式,