Factoring a polynomial in short and simple way.

Question

Question: I have been given a polynomial by my teacher. I have to tell him all real and complex factors of the following polynomial.

$(l+m+n)^5-l^5-m^5-n^5$ for reals$(l,m,n)$

What I can observe: It is symmetric and homogeneous.

My approach: I know $4$ of the factors namely $5,(l+m),(m+n),(n+l)$ by using multinomial theorem and using the odd powers.

Problem: I am unable to find $2$ of the factors. I see only one way to do it and it is to expand the polynomial but I dont want to do it because it will be very long.
Is there any shorter method?

Answer 1

final detail, the quadratic factor $$ l^2 + m^2 + n^2 + mn + nl + lm $$ is irreducible over the reals and complexes. If you write out, using variables $x,y,z$ and complex coefficients $a,b,c,d,e,f$ the form $$ (ax+by+cz)(dx+ey+fz) $$ then carefully construct the Hessian matrix $H$ of second partial derivatives with respect to $x,y,z$ in order, the determinant come out to be zero. For that matter, ignoring the determinant, all we need to do is set the triple $(x,y,z)$ to the traditional cross product $(a,b,c) \times (d,e,f)$ to get a vector that is sent to the origin by the Hessian matrix, considered as a linear transformation. That is $H$ is singular. Oh, if $(a,b,c)$ and $(d,e,f)$ are actually parallel, then the rank of $H$ reduces to one.

Your Hessian comes out as $$ \left( \begin{array}{ccc} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array} \right) $$

In comparison, the Hessian of $$ x^2 + y^2 + z^2 - yz - zx - xy $$ comes out $$ \left( \begin{array}{ccc} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{array} \right) $$ so this time the form can be factored; first step is $$ \frac{1}{4} (2x-y-z)^2 + \frac{3}{4} (y-z)^2 \; \; . $$ The actual factoring becomes $$ \left( \frac{1}{2} (2x-y-z) + \frac{i \sqrt 3}{2} (y-z) \right) \left( \frac{1}{2} (2x-y-z) - \frac{i \sqrt 3}{2} (y-z) \right) $$ which can be written without fractions if we introduce the cube root of $1$ given by $\omega = \frac{-1 + i \sqrt 3}{2}$ It becomes $$ (x + y \omega + z \omega^2)(x + y \omega^2 + z \omega ) = x^2 + y^2 + z^2 - yz - zx - xy $$

Answer 2

I can see a couple more vectors with $l^5+m^5+n^5=0=l+m+n$ other than $(1,-1,0)$ and their multiples and permutations. These involve complex numbers (roots of unity) though....

Answer 3

Hint.

$$ (l + m + n)^5 - l^5 - m^5 - n^5 = 5(l+m)(l+n)(m+n)((l+m+n)^2-(lm+ln+mn)) $$

Answer 4

Given the known factors, the remaining factor must be symmetric, and homogeneous of degree $2$, hence must have the form $$a(l^2+m^2+n^2)+b(lm+mn+nl)$$ for some constants $a,b$.

To find $a,b$, just choose some values for $l,m,n$, and substitute them into the equation $$(l+m+n)^5-l^5-m^5-n^5=5(l+m)(m+n)(n+l)\bigl(a(l^2+m^2+n^2)+b(lm+mn+nl)\bigr)$$ Using $(l,m,n)=(0,1,1)$ we get $$30=(10)(2a+b)$$ hence $2a+b=3$.

Using $(l,m,n)=(0,1,2)$ we get $$210=(30)(5a+2b)$$ hence $5a+2b=7$.

Solving the system \begin{align*} 2a+b&=3\\[4pt] 5a+2b&=7\\[4pt] \end{align*} yields $a=b=1$, hence the remaining factor is $$l^2+m^2+n^2+lm+mn+nl$$ To see that the above polynomial can't be factored further, substitute $l=1$.

Regarding the resulting polynomial $$m^2+mn+n^2+m+n+1$$ as a quadratic polynomial in $m$, the discriminant is $$-3n^2-2n-3$$ which has two distinct roots, so is not the square of a polynomial in $\mathbb{C}[n]$.

It follows that $l^2+m^2+n^2+lm+mn+nl$ is irreducible in $\mathbb{C}[m,n,l]$, hence we get the complete factorization $$(l+m+n)^5-l^5-m^5-n^5=5(l+m)(m+n)(n+l)(l^2+m^2+n^2+lm+mn+nl)$$