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

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!

• I suggest that you have a look at Kronecker's method. Jan 2, 2016 at 18:44
• Computers are much better at this sort of factoring than human beings are. We should probably just give up, and focus on things that we are good at, and let the machines do the grunt-work for us. Jan 3, 2016 at 3:58

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.

• And a Happy New Year to you too @Manu Jan 2, 2016 at 22:51

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.

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.

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.

• Why the downvote? Jan 2, 2016 at 18:36
• I didn't down vote you, but I suspect it was because whoever did thinks that your post wasn't an answer, but a vague suggestion of various techniques. Personally, I think your post makes a better comment. Jan 2, 2016 at 18:38
• "There is no answer" is an answer though. Jan 2, 2016 at 18:38
• Agreed, but I'm only speculating what the downvoter thought. I do believe that without more elaboration, your post makes a better comment than answer. Jan 2, 2016 at 18:41
• I didn't downvote it either, but your answer does not address OP's question. He isn't asking about a random polynomial, he is asking about a specific one. For instance, $x^7-1=(x-1)(\text{you know the rest})$ and I didn't need a computer software for that. Jan 2, 2016 at 18:45

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.

• Ha, when you see $-1+\sqrt2=0.414\dots$ simply because you know the expansion of $\sqrt2$ quite a few decimals? Oct 7, 2016 at 18:35
• Yeah it's handy memorizing the first few square root of the primes. @SimpleArt Oct 7, 2016 at 21:23
• To a few decimals. Oct 8, 2016 at 2:14