How to factor $x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1$ by hand?

Question

I know that

$x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1 = (x^4-x^3+x+1)(x^2+2x-1)$

but I would not know how to do that factoring without a software.

Some idea? Thank you!

Answer 1

The equation is palindromic (well, almost), so:

We can write it as $$x^3\left[x^3+x^2-3x+2+\frac 3x+\frac{1}{x^2}-\frac{1}{x^3}\right]$$ $$=x^3\left[(x^3-3x+\frac 3x-\frac{1}{x^3})+\left((x-\frac 1x)^2+2\right)+2\right]$$ $$=x^3\left[u^3+u^2+4\right],$$ where $u=x-\frac 1x$ And hence the factorization is $$x^3(u+2)(u^2-u+2)$$ which will give us the expected answer.

Answer 2

This isn't doable. Given a random polynomial of high degree and no other information, there isn't an easy way to factor the polynomial without computer assistance. Sometimes you can make progress using Galois Theory and looking at properties of extensions, but that doesn't always give you enough information either.

For polynomials of degree larger than $4$, factoring is not a task that can realistically be done by hand. In the cases of degree less than $5$, there is an equation you can use to find the rooms, and then reverse engineer the factors. In the degree 5 or 6 case it's possible to brute force it, but the computation is long and difficult to do without a computer (and requires calculating the inverse of a matrix of size $\deg(f)$). To solve it by brute force, you write out all the possible factors with generic coefficients and then try to solve the resulting system of equations.

Answer 3

A possibility is to write a generic product

$x^6+x^5-3x^4+2x^3+3x^2+x-1=(x^4+a_3x^3+a_2x^2+a_1x+a_0)(x^2+b_1x+b_0)$

then expand the right-hand side and compare the two polynomials, obtaining

\begin{cases} &a_0 b_0=-1,\\ &a_0 b_1+a_1 b_0=1,\\ &a_0+a_1 b_1+a_2 b_0=3,\\ &a_1+a_2 b_1+a_3 b_0=2,\\ &a_2+a_3 b_1+b_0=-3,\\ &a_3+b_1=1 \end{cases}

that can be solved rather easily for integer solutions.
The same could be tried for a product of two third degree polynomials, without any (integer) solutions.

Answer 4

The rational root theorem turns out to be of no use in this case.

Using a Non-Graphical Calculator

I would first start by using Newtons method to find some approximate roots and factor probably using synthetic division. Using a calculator and newtons method I got $x=0.414...$ to be one answer, which then I suspect to be $x=-1+\sqrt{2}$ and check to be true. And by the irrational root theorem I know another root is $x=-1-\sqrt{2}$. Multiplying out the two factors $(x+1+\sqrt{2})(x+1-\sqrt{2})$ gets you another factor of $x^2+2x-1$. From here it's just long division. Take note that we got really lucky however because we were able to recognize our approximation as a familiar number, usually we will just be stuck with approximations. But approximations can be quit helpful especially when you get two approximate roots that differ by an approximate rational number.

Answer 5

Note that $(x - 1/x)^{2k}$ gives $x^{2k} + x^{-2k}$ and other symmetric terms, and $(x - 1/x)^{2 k + 1} = x^{2 k + 1} - x^{- 2 k - 1}$ and other terms, it looks like due to the symmetry of the coefficients by dividing by $x^3$ you can reduce the degree to a cubic in $x - 1/x$, and go from there.