Constant craving has always been

Since $f(x)$ is a polynomial, it will have the form:

$a_ix^i + a_{i-1}x^{i-1} + \cdots + a_2x^2 + a_1x + a_0$

Realize that any component of the form $a_ix^i$ will be a multiple of the value of x . A multiple is the product of an integer and a quantity. So, whether we multiply x by itself through exponentiation or by a coefficient, we will always be able to divide the product by x .

Furthermore, the sum of multiples of x is also a multiple of x , since we are able to factor an x and divide it out. Therefore, the following segment of the polynomial is also a multiple of x :

$a_ix^i + a_{i-1}x^{i-1} + \cdots + a_2x^2 + a_1x$

Now, the only thing that we need to look at is the constant in the polynomial. As we saw in the previous statement, the sum is a multiple of x if the addends are also a multiple of x . However, this would mean the constant would have to be different for each case of n . This violates the definition of a constant.

The only constant that would not have an effect on the multiplicity of the rest of the polynomial would be $\fbox{0}$

Let $f(n) = a_in^i + a_{i-1}n^{i-1} + \ldots + a_1n + a_0.$ Notice that all of the terms in this expression are multiples of $n,$ except for $a_0,$ the constant term, which means that $a_0$ must be a multiple of $n.$ Therefore, $a_0$ must be a multiple of all positive integers. The only number with this property is $a_0 = \boxed0.$

It's simple. If the function $f(n)$ is divisible by all values of $n$ , then all the terms of the function must contain $n$ as a factor. But since the constant term always holds a constant value, the constant term, in this case, must be either $n$ or $0$ . Since it can be $n$ because it's value is not fixed, it is sure to be $0$ .

Let f(x)=a $x^{2}$ +bx+c.If f(x) is divisible by x,c must also be divisible by x.Thus c must be divisible by all positive values.0 is the only value which satisfies this.Thus 0 is the answer.

One can just let f(x)=x... What a silly problem.

Constant term is f(0) which is a multiple of 0 and all multiples of zero are zero so f(0)=0

f(x) = (x + a)^e

If a > 0 then f(n) cannot be divided evenly into n parts.

For example if f(x) = (x + a)^3 = x^3 + 3a^2 x + 3ax^2 + a^3

then f(x) / x = x^2 + 3a^2 + 3ax + a^3/x

So if a > 0 then the a^e term will always be a^e/x

If a = 0 then f(x) = x^e which can be divided by an x term only

f(x) / x = x^e / x = x^(e - 1)

f(x) = x^n + x^(n-1) + x + c

c = 0

so that

f(x) = x^n + x^(n-1) + x which is divisible by x itself

giving x^n ...+ c