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I generated this polynomial after playing around with the golden ratio. I first observed that (using various properties of $\phi$), $\phi^3+\phi^{-3}=4\phi-2$. This equation has no significance at all, I just mention it because the whole problem stems from me wondering: which other numbers does this equation hold for?

The six possible answers are the roots of $x^6-4x^4+2x^3+1=0$. Note that I am not interested in solving for $x$ itself as much as I am interested in a method which would allow me to completely factor out this polynomial into lowest degree factors which still have real coefficients. Note that I am treating this equation as if I had no clue that the golden ratio is one of the solutions. In other words, I am trying to factor this equation as if I never saw it before, so I can't just immediately factor out $(x^2-x-1)$ without a justifiable process, even though it is indeed one of the factors.

I first observed that the equation holds for $x=1$, so I was able to divide out $(x-1)$ to get the factorization of:

$$(x-1)(x^5+x^4-3x^3-x^2-x-1)$$

I tried making an assumption that the quintic reduces to a product of $(x^3+Ax^2+Bx+C)(x^2+Dx+E)$, multiplying out, and equalling coefficients, but I ended up with a system of two extremely convoluted equations which I had no idea how to solve. I also tried to turn the first five terms of the quintic into a palindromic polynomial and then perform the standard method of factoring palindromic polynomials, to no avail.

I am either missing something, or I don't know of a nice method that would let this expression be factored. I'm looking forward to being enlightened, thanks for any help.

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  • $\begingroup$ Are you familiar with long division? en.wikipedia.org/wiki/Polynomial_long_division $\endgroup$ Mar 1 '20 at 5:34
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    $\begingroup$ As the answer below suggests, Wolfram Alpha agrees with your approach. What was the problem with the system of equations? Also, that factorization does not seem to have real roots except $x=1$ $\endgroup$
    – gt6989b
    Mar 1 '20 at 5:34
  • $\begingroup$ @gt6989b I was able to reduce the resulting system of five equations into two equations by elimination, but the remaining two equations had a lot of terms like $AB^2, B^3, A^2, AB,$ etc.. in other words, I was helpless, unless I did it wrong somehow $\endgroup$
    – KKZiomek
    Mar 1 '20 at 5:37
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    $\begingroup$ If you use $x=10$, you get $962001=9\times 89\times 1201$ which should tell you something about the factorization $\endgroup$
    – Edward H
    Mar 1 '20 at 5:39
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    $\begingroup$ Due to Abel's impossibility theorem, you don't have an algorithm to exactly factorize a generic polynomial of degree at least $5$. The best you could do is probably using numerical approximations. You have to know some extra information about your polynomial if you want to factorize it exactly. So after dividing your degree-$6$ polynomial by $x-1$, you get a quintic polynomial, and unless you know some extra information, I don't think you can factorize it exactly. $\endgroup$ Mar 1 '20 at 5:44
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Here's a possible way to do it:

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

$(x^3+1)^2-4x^4 = [x^3+1-2x^2][x^3+1+2x^2]$

$x^6-4x^4+2x^3+1= [(x^3-x^2)+(1-x^2)][x^3+2x^2+1]$

Then, we have:

$x^6-4x^4+2x^3+1 =[x^3+2x^2+1][x^2(x-1)+(1-x)(1+x)]$

$x^6-4x^4+2x^3+1 = (x-1)(x^2-x-1)[x^3+2x^2+1]$

So that gives you a decently nice factored form.

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    $\begingroup$ That's exactly the type of answer I was looking for. Thank you! I have been at the step of the difference of squares at some point, but I couldn't break the cubics down. Your separation of $x^3-2x^2+1$ into $(x^3-x^2)+(1-x^2)$ was very clever! $\endgroup$
    – KKZiomek
    Mar 1 '20 at 5:53
  • $\begingroup$ Thank you :) I try to be clever (failing in most of my attempts, of course!) $\endgroup$
    – Abhi
    Mar 1 '20 at 5:54
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Your original method is tedious but it can be done.

You can show that $(x^3+Ax^2+Bx+C)(x^2+Dx+E)$ is equal to:

$$x^5+(D+A)x^4+(1+AD+B)x^3 + (AE+BD+C)x^2 + (BE+CD) + CE$$

so $A+D = 1, B+AD+1 = -3, AE+BD+C=-1, BE+CD=-1, CE=-1$.

Assuming $A,B,C,D,E$ are all integers, we either have $C=-1, E=1$ or $C=1, E=-1$.

If $C=-1, E=1$, then we have:

$$A+D=1 \tag{1}$$ $$B+AD=-4 \tag{2}$$ $$A+BD=0 \tag{3}$$ $$B-D=-1 \tag{4}$$

$(1)+(4)$ gives $A+B=0$ so $A=-B$, which gives:

$$-B+D=1 \tag{5}$$ $$B-BD=-4 \tag{6}$$ $$-B+BD=0 \tag{7}$$ $$B-D=-1 \tag{8}$$

and this is clearly impossible since $(6) + (7)$ gives $0=-4$.

Therefore we must have $C=1, E=-1$:

$$A+D=1 \tag{9}$$ $$B+AD=-4 \tag{10}$$ $$-A+BD=-2 \tag{11}$$ $$-B+D=-1 \tag{12}$$

This time $(9)-(12)$ gives $A+B=2$, so $A=2-B$:

$$-B+D=-1 \tag{13}$$ $$B+2D-BD=-4 \tag{14}$$ $$B+BD=0 \tag{15}$$ $$-B+D=-1 \tag{16}$$

$(14)+(15)$ gives $2B+2D = -4$, so $B+D=-2$. When we add this to $(16)$, $2D=-2$ so $D=-1$.

And the rest follows:

$$B - D = 1 \Rightarrow B+1=1, B=0$$ $$A=2-B \Rightarrow A=2$$

so the factorisation is $(x-1)(x^3+2x^2+1)(x^2-x-1)$.

I wouldn't wish this method on anybody.

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