[Math] Factor $x^4-7x^2+1$

factoringpolynomials

Is there a general method of factoring fourth order polynomials into a product of two irreducible quadratics?

As I am reviewing on finding roots of polynomials in $\mathbb Z_n$ for abstract algebra, I am trying to factor the polynomial $x^4-7x^2+1$, and I was given the answer of $(x^2+3x+1)(x^2-3x+1)$.

I was able to verify this of course by multiplying together the two irreducible quadratics, but I need to pretend that I never received the answer in the first place and ask for any hint in proceeding how to factor the fourth-order polynomial. Thanks.

It feels like I should use a method of "difference of squares" more than anything…

Best Answer

My first instinct in factoring is completing the square. $x^4-7x^2 + 1$ naturally becomes $(x^2-1)^2-5x^2$ or $(x^2+1)^2-9x^2$. Of the two the latter is a difference of squares and is thus more useful. Thus $$x^4-7x^2 + 1 = (x^2+1)^2-9x^2 = (x^2 + 1 + 3x)(x^2+1-3x)$$

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I might be misunderstanding, but this just looks like concealed expanding $$(x^2+ax+b)(x^2+cx+d)\tag{1}$$ and comparing coefficients.

Factoring extremes

Well, what we want is to choose $b$ and $d$ such that $(x^2+b)(x^2+d)$ matches leading and constant factor. What it amounts to is finding $b$ and $d$ such that $bd = -3$. Let's say we were lucky and chose the winning pair $b = -1,\ d=3.$

Calculating $\Delta$

Why is this important? Well, $(x^2-1)(x^2+3) = x^4+2x^2-3$, so we are missing one $x^2$: $\Delta = 3x^2 - 2x^2$.

Covering for the missing $\Delta$

There is only one place where $\Delta$ can come from, that is from $a$ and $c$ - we must have $ac x^2 = \Delta$. Why? Well, if you expand $(1)$, you will find that $x^2$ term must be equal to $b + d + ac$. We found $b+d$ in step 1 and 2, so $acx^2$ is equal to $\Delta$. In this case $ac = 1$. There are two possibilities in $\Bbb Z$, either $a=c=1$, or $a=c=-1$. It is easy to decide which one, since it must fit the other terms as well, namely $x^3$ term is equal to $a + c$ and the $x$ term is equal to $ad + bc = 3a - c$. Thus, $a = c = 1$.

Collecting terms

We simply write $(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3$ and feel the magic all around us.

Now, let us do the same thing, only transparently:

$$(x^2+ax+b)(x^2+cx+d) = x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$

so we need to solve the following system in $\Bbb Z$:

\begin{array}{c r}\begin{align} a+c&=2\\ b+d+ac&=3\\ ad+bc&=2\\ bd &= -3 \end{align} &\tag{2}\end{array}

There are two possibilities, either $b = 1$ and $d = -3$, or $b = -1$ and $d = 3$. We can see that former leads to no solution, so we will pursuit the latter, the system becomes:

\begin{align} a+c&=2\\ ac&=1\\ 3a-c&=2 \end{align}

The first and the third equation form linear system, which has unique solution $a = c = 1$, luckily, consistent with the second equation. Thus, we now know for certain that $$(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3.$$

Addendum:

There is a legitimate question why we would consider $(1)$ in the first case. Now, we are trying to factor a quartic over $\Bbb Z$. Let's say $p(x) = x^4 +2x^3+3x^2+2x-3$ and assume that $p = fg$. We have that $\deg p = \deg f + \deg g$, so there are two cases: $4=1+3$ or $4=2+2$. The first case would imply that $p$ has integer root and we can use rational root theorem to rule that out. Thus, only case $4=2+2$ remains, which leads to considering $(1)$.

Finally, if the system $(2)$ had no solutions in $\Bbb Z$, we would conclude that $p$ is irreducible over $\Bbb Z$.

Factor, or decompose, a polynomial related to the Heptadecagon

Here is how i would proceed. There will be some human calculation routine too, but the "technique" is a combined human+machine activity. The structure will come in a second, but so far, let us experiment something.

First of all, i asked sage about the Galois group of the number field generated by a root of the polynomial $$ P = x^8 + x^7 - 7x^6 - 6x^5 + 15x^4 + 10x^3 - 10x^2 - 4x + 1 $$ to be factorized. It is a cyclic group of order eight. I also asked for the subfields, and sage mentions a subfield of degree four over $\Bbb Q$, related to the polynomial $$ Q = x^4 + x^3 - 6 x^2 - x + 1 \ . $$ (Code and results are postponed to not disturb here the flow. A structural way to see it follows below.) Let us compute its roots. This is a "twisted reciprocal" polynomial, one can substitute $X=ix$ first to get a reciprocal polynomial. The substitution $$ t = x-\frac 1x $$ allows to relate $Q/x^2$ with $$ T = t^2 + t -4\ . $$ (This polynomial is also present in the sage list of subfields...) The roots of $T$ are $t_{\pm}=(-1\pm \sqrt{17})/2$. The polynomials $x^2-t_\pm x-1$ have roots $$ \begin{aligned} x_{\boxed\pm} &=\frac 12\left(t_\pm\boxed\pm\sqrt{t_\pm^2+4}\right) \\ &=\frac 12\left(t_\pm\boxed\pm\sqrt{8-t_\pm}\right) \\ &= \frac 14\left(-1\pm\sqrt{17}\right) \boxed\pm \frac 12\sqrt{\frac 12\left(17\mp\sqrt{17}\right)} \ . \\ 4x_{\boxed\pm} &= \left(-1\pm\sqrt{17}\right) \boxed\pm \sqrt{2\left(17\mp\sqrt{17}\right)} \ . \end{aligned} $$

Time for the structure. I learned it from an older book of Hans Rademacher, typed it for myself some years ago. The idea is to use Gaussian periods. Let $K=\Bbb Q(w)$ be the cyclotomic field generated by the primitive root of order $17$, here fixed and denoted by $w$ instead of the longer $\zeta_{17}$, which is hard to type. The Galois group of $K$ over $\Bbb Q$ has elements sending $w$ to some other primitive root $w^k$, so $k\in 1,2,\dots 16$.

My choice is $3$ as a generator of the cyclic multiplicatie group $(\Bbb Z/17)^\times$, so let us use the Frobenius morphism $$ Fw = w^3\ .$$

So $F^2w = FFw=Fw^3=(w^3)^3=w^9$, and $F^3w = FF^2w=Fw^9=(w^3)^9 = w^{27}$, and in general $F^jw = w^{3^j}$.

The idea of the Gaussian periods is simple. We associate the following sums: $$ \begin{aligned} S &= F^0w + Fw + F^2w + F^3w+\dots+F^{14}w+F^{15}w\\ &=w+w^2+w^3+\dots+w^{16}\text{ (in other order)}\\ &=-1\ ,\\[2mm] &\qquad\text{ and we split it in two subsums,}\\[2mm] S_0 &= F^0w + F^2w + F^4w+\dots+F^{14}w\\ S_1 &= Fw + F^3w + F^5w+\dots+F^{15}w\\[2mm] &\qquad\text{ and we split each in two subsums,}\\[2mm] S_{00} &= F^0w + F^4w + F^8w+F^{12}w\\ S_{10} &= F^2w + F^6w + F^{10}w+F^{14}w\\[2mm] S_{01} &= Fw + F^5w + F^9w+\dots+F^{13}w\\ S_{11} &= F^3w + F^7w + F^{11}w+\dots+F^{15}w\\[2mm] &\qquad\text{ and we split each in two subsums,}\\[2mm] S_{000}&= F^0w + F^8w\\ S_{100}&= F^4w + F^{12}w\\ &\qquad\text{and so on...} \end{aligned} $$ The indices are my choice, they suggest binary written numbers. So $S_{01}$ is a sum of terms $F^jw$ with $j$ congruent to (the binary phone number) $01_{[2]}=1$ modulo $4$, $S_{101}=F^5w+F^{13}w$ and so on.

There are obvious splitting relations, for instance $S_0+S_1=S=-1$, $S_{00}+S_{10}=S_0$, sum relations. But there are also product relations.

Instead of writing them here explicitly, a hard jax job, i will use sage to "type" them. Sage is a mathematically structurally thinking language, so i hope the dry exposition gives the information in a human way.

sage: K.<w> = CyclotomicField(17)
sage: K
Cyclotomic Field of order 17 and degree 16
sage: w^17
1

sage: S     = sum( [ w^(3^j) for j in [0..15] ] ) 

sage: S0    = sum( [ w^(3^j) for j in [0..15] if j%2 == 0 ] ) 
sage: S1    = sum( [ w^(3^j) for j in [0..15] if j%2 == 1 ] ) 

sage: S00   = sum( [ w^(3^j) for j in [0..15] if j%4 == 0 ] ) 
sage: S01   = sum( [ w^(3^j) for j in [0..15] if j%4 == 1 ] ) 
sage: S10   = sum( [ w^(3^j) for j in [0..15] if j%4 == 2 ] )
sage: S11   = sum( [ w^(3^j) for j in [0..15] if j%4 == 3 ] )

sage: S000  = sum( [ w^(3^j) for j in [0..15] if j%8 == 0 ] ) 
sage: S001  = sum( [ w^(3^j) for j in [0..15] if j%8 == 1 ] )
sage: S010  = sum( [ w^(3^j) for j in [0..15] if j%8 == 2 ] )
sage: S011  = sum( [ w^(3^j) for j in [0..15] if j%8 == 3 ] )
sage: S100  = sum( [ w^(3^j) for j in [0..15] if j%8 == 4 ] )
sage: S101  = sum( [ w^(3^j) for j in [0..15] if j%8 == 5 ] )
sage: S110  = sum( [ w^(3^j) for j in [0..15] if j%8 == 6 ] )
sage: S111  = sum( [ w^(3^j) for j in [0..15] if j%8 == 7 ] )

sage: S0+S1, S0*S1
(-1, -4)
sage: S0.minpoly()
x^2 + x - 4

sage: S00*S10, S01*S11
(-1, -1)
sage: S00.minpoly()
x^4 + x^3 - 6*x^2 - x + 1
sage: S10.minpoly()
x^4 + x^3 - 6*x^2 - x + 1
sage: S01.minpoly()
x^4 + x^3 - 6*x^2 - x + 1
sage: S11.minpoly()
x^4 + x^3 - 6*x^2 - x + 1

sage: S000*S100
w^14 + w^12 + w^5 + w^3

sage: S00, S10, S01, S11
(-w^15 - w^14 - w^12 - w^11 - w^10 - w^9 - w^8 - w^7 - w^6 - w^5 - w^3 - w^2 - 1,
 w^15 + w^9 + w^8 + w^2,
 w^14 + w^12 + w^5 + w^3,
 w^11 + w^10 + w^7 + w^6)
sage: S000*S100 == S01
True

(The first long expression staring with -w^15 - w^14 ... is only the computer version to hide the "missing terms" of the minimal polynomial of $w$, the four that not appear in the list, since it thinks it is a good idea to not use $w^{16}$...)

Note that in the "naive approach" from the beginning was also dealing with polynomials that appear as minimal polynomials of the one or the other $S_?$.

The OP wants to understand the splitting of the polynomial $P$, which is the minimal polynomial (over rationals) of $$S_{000}=w+w^{16}=w+\frac 1w\ .$$

And now we can give four factors of $P$. With the computer aid of sage:

sage: R.<x> = PolynomialRing(K)
sage: R
Univariate Polynomial Ring in x over Cyclotomic Field of order 17 and degree 16
sage: (x-S000)*(x-S100)
x^2 + (w^15 + w^14 + w^12 + w^11 + w^10 + w^9 + w^8 + w^7 + w^6 + w^5 + w^3 + w^2 + 1)*x + w^14 + w^12 + w^5 + w^3
sage: (x-S000)*(x-S100) == x^2 - S00*x + S01
True
sage: (x-S010)*(x-S110)
x^2 + (-w^15 - w^9 - w^8 - w^2)*x + w^11 + w^10 + w^7 + w^6
sage: (x-S010)*(x-S110) == x^2 - S10*x + S11
True
sage: (x-S001)*(x-S101)
x^2 + (-w^14 - w^12 - w^5 - w^3)*x + w^15 + w^9 + w^8 + w^2
sage: (x-S001)*(x-S101) == x^2 - S01*x + S10
True
sage: (x-S011)*(x-S111)
x^2 + (-w^11 - w^10 - w^7 - w^6)*x - w^15 - w^14 - w^12 - w^11 - w^10 - w^9 - w^8 - w^7 - w^6 - w^5 - w^3 - w^2 - 1
sage: (x-S011)*(x-S111) == x^2 - S11*x + S00
True

sage: (x-S00)*(x-S10)*(x-S01)*(x-S11)
x^4 + x^3 - 6*x^2 - x + 1

And here explicitly, because it is structural (the "traces", coefficients in $x$ on the RHSs are easy, we splitted them as the idea to proceed, for the computation of the free coefficients there is also a Galois pattern...): $$ \begin{aligned} (x-S_{000})(x-S_{100}) &= x^2 - S_{00}x+ S_{01}\ ,\\ (x-S_{010})(x-S_{110}) &= x^2 - S_{10}x+ S_{11}\ ,\\ (x-S_{001})(x-S_{101}) &= x^2 - S_{01}x+ S_{10}\ ,\\ (x-S_{011})(x-S_{111}) &= x^2 - S_{11}x+ S_{00}\ ,\\[2mm] &\text{where}\\ (x-S_{00}) (x-S_{10}) (x-S_{01}) (x-S_{11}) &=x^4 + x^3 - 6x^2 - x + 1\ , \end{aligned} $$ and its roots were computed in advance.

I have to submit, but there remains a lot to say, maybe just mention that this structure made the 19 years old Gauss to choose mathematics, not physics, one main reason for still missing a unification of matter in this field...

Postponed: The sage code used to obtain the polynomial $Q$, the Galois theoretical information, and the decomposition of $P$.

sage: R.<x> = PolynomialRing( QQ )
sage: P = x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
sage: K.<v> = NumberField( P )
sage: v.minpoly()
x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
sage: G = K.galois_group()
sage: G
Galois group of Number Field in v with defining polynomial x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
sage: G.is_cyclic()
True
sage: K.subfields()
[
(Number Field in v0 with defining polynomial x + 1, Ring morphism:
  From: Number Field in v0 with defining polynomial x + 1
  To:   Number Field in v with defining polynomial x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
  Defn: -1 |--> -1, None),
(Number Field in v1 with defining polynomial x^2 + x - 4, Ring morphism:
  From: Number Field in v1 with defining polynomial x^2 + x - 4
  To:   Number Field in v with defining polynomial x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
  Defn: v1 |--> -v^7 - v^6 + 6*v^5 + 6*v^4 - 10*v^3 - 9*v^2 + 5*v + 1, None),
(Number Field in v2 with defining polynomial x^4 + x^3 - 6*x^2 - x + 1, Ring morphism:
  From: Number Field in v2 with defining polynomial x^4 + x^3 - 6*x^2 - x + 1
  To:   Number Field in v with defining polynomial x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
  Defn: v2 |--> v^4 - 4*v^2 + v + 2, None),
(Number Field in v3 with defining polynomial x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1, Ring morphism:
  From: Number Field in v3 with defining polynomial x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
  To:   Number Field in v with defining polynomial x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
  Defn: v3 |--> v, Ring morphism:
  From: Number Field in v with defining polynomial x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
  To:   Number Field in v3 with defining polynomial x^8 + x^7 - 7*x^6 - 6*x^5 + 15*x^4 + 10*x^3 - 10*x^2 - 4*x + 1
  Defn: v |--> v3)
]

sage: K2, morphism, _ = K.subfields()[2]
sage: K2
Number Field in v2 with defining polynomial x^4 + x^3 - 6*x^2 - x + 1
sage: K2.gen()
v2
sage: for f, mul in P.base_extend(K2).factor():
....:     print f
....:     
x^2 - v2*x - 1/2*v2^3 + 3*v2 - 3/2
x^2 + (-v2^3 - v2^2 + 6*v2 + 1)*x - 1/2*v2^3 - v2^2 + 2*v2 + 3/2
x^2 + (1/2*v2^3 - 3*v2 + 3/2)*x + v2^3 + v2^2 - 6*v2 - 1
x^2 + (1/2*v2^3 + v2^2 - 2*v2 - 3/2)*x + v2

Best Answer

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