Quaint Quintic

Say that we have some polynomial $f(x)=c_kx^{k}+c_{k-1}x^{k-1}+\cdots+c_0$ . And, our assumption is: $\sum_{n=0}^{k}c_n=0$

Remember that our polynomial is of the form: $f(x)=\sum_{n=0}^{k}c_nx^{n}$ Letting $x=1$ , we then have: $f(1)=\sum_{n=0}^{k} c_n1^{n-1}=\sum_{n=0}^{k}c_n=0$ By our assumption. Hence $1$ is an integer root of the polynomial.

Sum of coefficients=f(1); Thus f(1)=0;Thus 1 is the integer root.

$f(x)=ax^5+bx^4+cx^3+dx^2+ex$ is such a polynomial. So, $a+b+c+d+e=0$

We can conclude that, considering $x$ with powers are now co-efficients of $a,b,c,d,e$ , we get, $x=1$ is the root.

Describing in the "if and only if" way: $f(1)=a+b+c+d+e=0$ , so $1$ is the root.

Again, if $1$ is a root, then, $f(1)=a+b+c+d+e=0$ , which makes, the answer is $1$ .

$f(x)$ is a quantic polynomial with only one integer root... Hence it's a perfect fifth power of an integer... We can write $f(x)$ as: $f(x)=\left(x-a\right)^5$ where $a$ is the integer root... Using binomial expansion formula, we can expand $(x-a)^5$ and get...

$f(x)=x^5-5ax^4+10a^2x^3-10a^3x^2+5a^4x-a^5$ ... Since, the sum of it's coefficients is $0$ , we can form an equation...

$1 - 5a + 10a^2-10a^3+5a^4-a^5=0$ ... It's not quite tough to find out the root $a=1$ ...

Hence, the integer root is: $\fbox{1}$ ...

Remark: The given polynomial $f$ is $f(x) = \left(x-1\right)^2 = x^5-5x^4+10x^3-10x^2+5x-1$

Well, obviously

$f(x) = \sum_{n=0}^{5} a_{n}x^{n}$

And then

$f(1) = \sum_{n=0}^{5} a_{n} = 0$

Then $x = 1$ is a root.

Our polynomial is in the form $(x-n)^5$ , where $n$ is our desired root. This is because our polynomial only has one root, so it must be in this form.

The sum of the coefficients of $(x-n)^5$ can be found by simplifying the whole thing, but there is a much simpler method.

For example, if we want the sum of the coefficients of $(x^2-4)(x+9)$ , we can just get rid of the $x$ terms, leave the coefficients, and multiply:

$(1-4)(1+9) =-3\cdot10=-30$ .

Similarly, we want the sum of the coefficients of $(x-n)^5$ to be equal to 0. Since the $x$ term has a coefficient of 1, the sum of the coefficients is simply $(1-n)^5$ . If we want this to be equal to 0, we know that $1-n$ must be equal to 0, which leads us to $n=1$ . This means that the integer root (which is equal to $n$ ) is equal to $\boxed{1}$ .

(Note: the polynomial is actually in the form $(ax-n)^5$ , because it is not specified that the $x$ coefficient is 1. However, this does not affect the answer: we will find (in a similar method as above) that the sum of the coefficients is equal to $(a-n)^5$ , and if we want this to be 0, then $a=n$ , which leads us to $n=1$ .)

Let $f$ be a quintic polynomial. By definition, $t$ is a root of $f$ if and only if $f(t)=0$ .

Here is the key step : the sum of the coefficients of $f$ is equal to $f(1)$ , indeed:

Let $a_0, a_1, ..., a_5$ be the coefficients of $f$ , $f(1)=a_5\times1^5+a_4\times1^4+a_3\times1^3+a_2\times1^2+a_1\times1+a_0=a_0+a_1+a_2+a_3+a_4+a_5$ which is the sum of the coefficients of $f$ .

The problem tells us that the sum of the coefficients of f is 0. So $f(1)=0$ . As a matter of fact, we can conclude, by the definition of a root, that $1$ is a root of the quintic polynomial $f$

If we take the general equation of a quintic polynomial.. we get sum of coefficient if x=1. When sum of coefficient is zero then x=1 is a root.

1 is always a solution of any polynomial whose coefficients add up to zero. So, answer is "1"

if F is 0. then the numerical value of x is 1. so f sub x. = 1

The answer is "1" because only "1" has the one integer root, since the coefficients of f(x) is "0", So if you add +1 and -1 you get zero.

as the 5 degree has only a root it is an expantiion of (x-a)^{5} where a is its root but 2^{nd} cond^{n} gives us the root to be 1 since the sum of all coefficients of a polyomial which is expansion of a binomial term is 0 only when it is expansion of (x - 1)^{n} i.e. its root is 1.