- #1
juantheron
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Find the number of polynomials of degree $5$ with distinct coefficients from the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$ that are divisible by $x^2 - x + 1$
chisigma said:If 'polynomials with distinct coefficients' means 'different polynomials' the solution is easy: the requested number is equal to the number of polynomials of degree 3 that can be constructed with a set of 8 coefficients, i.e. $\displaystyle n= 8^{4}= 4096$...
Kind regards
$\chi$ $\sigma$
Start by dividing the fifth degree polynomial $ax^5+bx^4+cx^3+dx^2+ex+f$ by $x^2-x+1$: $$ax^5+bx^4+cx^3+dx^2+ex+f = (x^2-x+1)\bigl(ax^3 + (a+b)x^2 + (b+c)x + (d+c-a)\bigr) + (d+e-b-a)x + (a+f-c-d).$$ The conditions for the remainder to be zero are $$\boxed{\begin{array}{c}a+b = d+e, \\ a+f = c+d. \end{array}}$$ So you are looking for the number of sextuplets $(a,b,c,d,e,f)$ consisting of six distinct elements from the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$ that satisfy the boxed conditions.jacks said:Find the number of polynomials of degree $5$ with distinct coefficients from the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$ that are divisible by $x^2 - x + 1$
jacks said:Find the number of polynomials of degree $5$ with distinct coefficients from the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$ that are divisible by $x^2 - x + 1$
chisigma said:Let's write the 5 degree polynomial as $\displaystyle P(x)= a_{5}\ x^{5} + a_{4}\ x^{4} + a_{3}\ x^{3} + a_{2}\ x^{2} + a_{1}\ x + a_{0}$. If $\displaystyle x^{2} - x +1$ divides P(x), then $\displaystyle e^{i\ \frac{\pi}{3}}$ as well as $\displaystyle e^{- i\ \frac{\pi}{3}}$ is a root of P(x), so that we can write...
$\displaystyle a_{5}\ e^{i\ \frac{5}{3}\ \pi} + a_{4}\ e^{i\ \frac{4}{3}\ \pi} + a_{3}\ e^{i\ \pi} + a_{2}\ e^{i\ \frac{2}{3}\ \pi} + a_{1}\ e^{i\ \frac{1}{3}\ \pi} + a_{0}= (a_{2}-a_{5})\ e^{i\ \frac{2}{3}\ \pi} + (a_{1}-a_{4})\ e^{i\ \frac{1}{3}\ \pi} + (a_{0}-a_{3})=0$ (1)
Observing (1) we note that all the triplets $\displaystyle a_{0},a_{1},a_{2}$ combined with the conditions $\displaystyle a_{3}=a_{0}, a_{4}=a_{1}, a_{5}=a_{2}$ satisfies (1) so that the requested number of polynomials is $\displaystyle 8^{3}= 512$...
If you take the real part of (1) then you find that $\frac12((a_{2}-a_{5}) +\frac12(a_{1}-a_{4}) + (a_{0}-a_{3}) = 0$. If you take the imaginary part of (1) then you get $a_{2}-a_{5} = a_{1}-a_{4}$. Those equations are equivalent to the boxed equations in my previous comment. These are the only conclusions that you can deduce from (1), and chisigma's following claim is incorrect:chisigma said:$\displaystyle a_{5}\ e^{i\ \frac{5}{3}\ \pi} + a_{4}\ e^{i\ \frac{4}{3}\ \pi} + a_{3}\ e^{i\ \pi} + a_{2}\ e^{i\ \frac{2}{3}\ \pi} + a_{1}\ e^{i\ \frac{1}{3}\ \pi} + a_{0}= (a_{2}-a_{5})\ e^{i\ \frac{2}{3}\ \pi} + (a_{1}-a_{4})\ e^{i\ \frac{1}{3}\ \pi} + (a_{0}-a_{3})=0$ (1)
chisigma said:Observing more carefully (1) we discover that the condition $\displaystyle a_{2}-a_{5}= a_{1}-a_{4}=a_{0}-a_{3}=0$ is not the only solving way because the basic equality $\displaystyle e^{i\ \frac{2}{3}\ \pi} - e^{i\ \frac{1}{3}\ \pi} + 1=0$ implies that $\displaystyle a_{2}-a_{5}= a_{4}-a_{1}=a_{0}-a_{3}= \pm 1$ is also a solving way and that means that the requested number of polynomials is $\displaystyle 8^{3} +2\ \cdot 7^{3}= 1198$...
I understood the phrase in red to mean that, in each such polynomial, the six integers $a_0,a_1,a_2,a_3,a_4,a_5$ should be distinct elements of the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$. That condition makes the problem very much harder to solve.jacks said:Find the number of polynomials of degree $5$ with distinct coefficients from the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$ that are divisible by $x^2 - x + 1$
Opalg said:If you take the real part of (1) then you find that $\frac12((a_{2}-a_{5}) +\frac12(a_{1}-a_{4}) + (a_{0}-a_{3}) = 0$. If you take the imaginary part of (1) then you get $a_{2}-a_{5} = a_{1}-a_{4}$. Those equations are equivalent to the boxed equations in my previous comment. These are the only conclusions that you can deduce from (1), and chisigma's following claim is incorrect:
The original problem was:
I understood the phrase in red to mean that, in each such polynomial, the six integers $a_0,a_1,a_2,a_3,a_4,a_5$ should be distinct elements of the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$. That condition makes the problem very much harder to solve.
>function CountDiv()
$
$ global I;
$
$ count=0;
$ OKs=0;
$ r1=0.5*(1+sqrt(3)*I);
$ r2=0.5*(1-sqrt(3)*I);
$ arry1=1:8;
$
$ for idx1=1 to 7
$ for idx2=idx1+1 to 8
$ aa=nonzeros(arry1!=idx1);
$ arry2=arry1(aa);
$ aa=nonzeros(arry2!=idx2);
$ arry2=arry2(aa);
$
$ for jdx=1 to fak(6)-1
$ count=count+1;
$ c1=polyval(arry2,r1);
$ if c1~=0
$ c2=polyval(arry2,r2);
$ if c2~=0
$ OKs=OKs+1;
$ endif
$ endif
$ arry2=next(arry2);
$ end
$ end
$ end
$
$ return {count, OKs}
$ endfunction
>{C,O}=CountDiv;[C,O]
20132 300
There are an infinite number of polynomials of degree 5 that are divisible by x2-x+1. This is because any polynomial of degree 5 can be written as (x2-x+1) multiplied by another polynomial of degree 3, resulting in a product of degree 5.
No, x2-x+1 is not a factor of all polynomials of degree 5. It is only a factor of those polynomials that are divisible by it.
x2-x+1 is a quadratic polynomial that has complex roots, meaning it cannot be factored into linear factors with real coefficients. This makes it a unique and interesting factor for polynomials of degree 5.
Yes, polynomials of degree 5 can be divisible by more than one polynomial. For example, a polynomial of degree 5 can be divisible by both x2-x+1 and x+2, resulting in a product of degree 7.
The number of polynomials of degree 5 divisible by x2-x+1 cannot be determined exactly, as it is infinite. However, it can be approximated by finding the number of polynomials of degree 5 with complex roots, which is also infinite.