Mixing
Factorise $8wxyz+w²yz+w²xz+w²xy+x²yz+x²wz+x²wy+y²xz+y²wz+y²wx+z²xy+z²wy+z²wx+w²xyz+wx²yz+wxy²z+wxyz²+xyz+wyz+wxz+wxy=0.$
Clash Royale CLAN TAG#URR8PPP
up vote
14
down vote
favorite
I apologize for not having anything to show for an attempt. I've provided additional context, though, to keep it on topic. I hope it's enough.
The Question:
Factorise $$beginalign, & 8wxyz \
&+w^2yz+w^2xz+w^2xy \
&+x^2yz+x^2wz+x^2wy \
&+y^2xz+y^2wz+y^2wx \
&+z^2xy+z^2wy+z^2wx \
&+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \
&+xyz+wyz+wxz+wxy=0tagI
endalign$$ over $Bbb C$.
Context:
Here's a Q&A based on an answer from this meta question:
What are you studying?
A PhD in combinatorial group theory, first year.
What text is this drawn from, if any? If not, how did the question arise?
None. It arose as part of my research into the orders of abelianisations of certain cyclically presented groups.
What kind of approaches (to similar problems) are you familiar with?
Here's a related question about a system of equations generated by a smaller case:
The system $2XY+X^2+Y^2+X^2Y+Y+XY^2+X=0$, for distinct pairs $X, YinJ, K, L$ for $n$th roots of unity $J, K, L$.
What kind of answer are you looking for? Basic approach, hint, explanation, something else?
A "simple" factorisation would suffice.
Is this question something you think should be able to answer? Why or why not?
No. I have little training in factorising polynomials in multiple variables.
Observation:
The equation $(I)$ has $w=x=y=z=0$ as a solution.
Please help :)
NB: I $colorredtextsuspect$ due to the context of the question that some of the roots are $n$th roots of unity. It would be weird if they weren't.
polynomials factoring
asked Jul 15 at 0:26
Shaun
7,41392972
 |Â
show 5 more comments
up vote
14
down vote
favorite
I apologize for not having anything to show for an attempt. I've provided additional context, though, to keep it on topic. I hope it's enough.
The Question:
Factorise $$beginalign, & 8wxyz \
&+w^2yz+w^2xz+w^2xy \
&+x^2yz+x^2wz+x^2wy \
&+y^2xz+y^2wz+y^2wx \
&+z^2xy+z^2wy+z^2wx \
&+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \
&+xyz+wyz+wxz+wxy=0tagI
endalign$$ over $Bbb C$.
Context:
Here's a Q&A based on an answer from this meta question:
What are you studying?
A PhD in combinatorial group theory, first year.
What text is this drawn from, if any? If not, how did the question arise?
None. It arose as part of my research into the orders of abelianisations of certain cyclically presented groups.
What kind of approaches (to similar problems) are you familiar with?
Here's a related question about a system of equations generated by a smaller case:
The system $2XY+X^2+Y^2+X^2Y+Y+XY^2+X=0$, for distinct pairs $X, YinJ, K, L$ for $n$th roots of unity $J, K, L$.
What kind of answer are you looking for? Basic approach, hint, explanation, something else?
A "simple" factorisation would suffice.
Is this question something you think should be able to answer? Why or why not?
No. I have little training in factorising polynomials in multiple variables.
Observation:
The equation $(I)$ has $w=x=y=z=0$ as a solution.
Please help :)
NB: I $colorredtextsuspect$ due to the context of the question that some of the roots are $n$th roots of unity. It would be weird if they weren't.
polynomials factoring
asked Jul 15 at 0:26
Shaun
7,41392972
It is not quite clear what field this is over, and whether there are maybe additional constraints. Taking the real case at face value, suppose that $,x=y=z=a ne 0,$ for example, then the equation reduces to the quadratic $,(a + 3) w^2 + (3 a^2 + 14 a + 3) w + 3 a^2 + a=0,$, which can certainly have non-zero $,w,$ roots of magnitude other than $1,$.
– dxiv
Jul 15 at 1:18
1
@dxiv I'll probably ask a separate question about a system of equations based on this, similar to the one I linked to in the question.
– Shaun
Jul 15 at 1:40
1
Horner Form: $z (x (w (1 + w) + w x) + w x z) + y (x (w (1 + w) + w x) + y (w x + (w + (1 + w) x) z) + z (w (1 + w) + x (1 + w (8 + w) + (1 + w) x) + (w + (1 + w) x) z))$
– Mason
Jul 15 at 1:41
1
There are numerous professional tools for factoring polynomials. Have you tried turning any of them loose on this?
– Steven Stadnicki
Jul 15 at 1:43
6
If it were 5wxyz instead of 8wxyz, there would be the factor 1+w+x+y+z
– Empy2
Jul 15 at 3:42
 |Â
show 5 more comments
up vote
14
down vote
favorite
up vote
14
down vote
favorite
I apologize for not having anything to show for an attempt. I've provided additional context, though, to keep it on topic. I hope it's enough.
The Question:
Factorise $$beginalign, & 8wxyz \
&+w^2yz+w^2xz+w^2xy \
&+x^2yz+x^2wz+x^2wy \
&+y^2xz+y^2wz+y^2wx \
&+z^2xy+z^2wy+z^2wx \
&+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \
&+xyz+wyz+wxz+wxy=0tagI
endalign$$ over $Bbb C$.
Context:
Here's a Q&A based on an answer from this meta question:
What are you studying?
A PhD in combinatorial group theory, first year.
What text is this drawn from, if any? If not, how did the question arise?
None. It arose as part of my research into the orders of abelianisations of certain cyclically presented groups.
What kind of approaches (to similar problems) are you familiar with?
Here's a related question about a system of equations generated by a smaller case:
The system $2XY+X^2+Y^2+X^2Y+Y+XY^2+X=0$, for distinct pairs $X, YinJ, K, L$ for $n$th roots of unity $J, K, L$.
What kind of answer are you looking for? Basic approach, hint, explanation, something else?
A "simple" factorisation would suffice.
Is this question something you think should be able to answer? Why or why not?
No. I have little training in factorising polynomials in multiple variables.
Observation:
The equation $(I)$ has $w=x=y=z=0$ as a solution.
Please help :)
NB: I $colorredtextsuspect$ due to the context of the question that some of the roots are $n$th roots of unity. It would be weird if they weren't.
polynomials factoring
asked Jul 15 at 0:26
Shaun
7,41392972
I apologize for not having anything to show for an attempt. I've provided additional context, though, to keep it on topic. I hope it's enough.
The Question:
Factorise $$beginalign, & 8wxyz \
&+w^2yz+w^2xz+w^2xy \
&+x^2yz+x^2wz+x^2wy \
&+y^2xz+y^2wz+y^2wx \
&+z^2xy+z^2wy+z^2wx \
&+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \
&+xyz+wyz+wxz+wxy=0tagI
endalign$$ over $Bbb C$.
Context:
Here's a Q&A based on an answer from this meta question:
What are you studying?
A PhD in combinatorial group theory, first year.
What text is this drawn from, if any? If not, how did the question arise?
None. It arose as part of my research into the orders of abelianisations of certain cyclically presented groups.
What kind of approaches (to similar problems) are you familiar with?
Here's a related question about a system of equations generated by a smaller case:
The system $2XY+X^2+Y^2+X^2Y+Y+XY^2+X=0$, for distinct pairs $X, YinJ, K, L$ for $n$th roots of unity $J, K, L$.
What kind of answer are you looking for? Basic approach, hint, explanation, something else?
A "simple" factorisation would suffice.
Is this question something you think should be able to answer? Why or why not?
No. I have little training in factorising polynomials in multiple variables.
Observation:
The equation $(I)$ has $w=x=y=z=0$ as a solution.
Please help :)
NB: I $colorredtextsuspect$ due to the context of the question that some of the roots are $n$th roots of unity. It would be weird if they weren't.
polynomials factoring
asked Jul 15 at 0:26
Shaun
7,41392972
edited Aug 11 at 12:13
asked Jul 15 at 0:26
Shaun
7,41392972
asked Jul 15 at 0:26
Shaun
7,41392972
asked Jul 15 at 0:26
Shaun
7,41392972
7,41392972
It is not quite clear what field this is over, and whether there are maybe additional constraints. Taking the real case at face value, suppose that $,x=y=z=a ne 0,$ for example, then the equation reduces to the quadratic $,(a + 3) w^2 + (3 a^2 + 14 a + 3) w + 3 a^2 + a=0,$, which can certainly have non-zero $,w,$ roots of magnitude other than $1,$.
– dxiv
Jul 15 at 1:18
1
@dxiv I'll probably ask a separate question about a system of equations based on this, similar to the one I linked to in the question.
– Shaun
Jul 15 at 1:40
1
Horner Form: $z (x (w (1 + w) + w x) + w x z) + y (x (w (1 + w) + w x) + y (w x + (w + (1 + w) x) z) + z (w (1 + w) + x (1 + w (8 + w) + (1 + w) x) + (w + (1 + w) x) z))$
– Mason
Jul 15 at 1:41
1
There are numerous professional tools for factoring polynomials. Have you tried turning any of them loose on this?
– Steven Stadnicki
Jul 15 at 1:43
6
If it were 5wxyz instead of 8wxyz, there would be the factor 1+w+x+y+z
– Empy2
Jul 15 at 3:42
 |Â
show 5 more comments
It is not quite clear what field this is over, and whether there are maybe additional constraints. Taking the real case at face value, suppose that $,x=y=z=a ne 0,$ for example, then the equation reduces to the quadratic $,(a + 3) w^2 + (3 a^2 + 14 a + 3) w + 3 a^2 + a=0,$, which can certainly have non-zero $,w,$ roots of magnitude other than $1,$.
– dxiv
Jul 15 at 1:18
1
@dxiv I'll probably ask a separate question about a system of equations based on this, similar to the one I linked to in the question.
– Shaun
Jul 15 at 1:40
1
Horner Form: $z (x (w (1 + w) + w x) + w x z) + y (x (w (1 + w) + w x) + y (w x + (w + (1 + w) x) z) + z (w (1 + w) + x (1 + w (8 + w) + (1 + w) x) + (w + (1 + w) x) z))$
– Mason
Jul 15 at 1:41
1
There are numerous professional tools for factoring polynomials. Have you tried turning any of them loose on this?
– Steven Stadnicki
Jul 15 at 1:43
6
If it were 5wxyz instead of 8wxyz, there would be the factor 1+w+x+y+z
– Empy2
Jul 15 at 3:42
It is not quite clear what field this is over, and whether there are maybe additional constraints. Taking the real case at face value, suppose that $,x=y=z=a ne 0,$ for example, then the equation reduces to the quadratic $,(a + 3) w^2 + (3 a^2 + 14 a + 3) w + 3 a^2 + a=0,$, which can certainly have non-zero $,w,$ roots of magnitude other than $1,$.
– dxiv
Jul 15 at 1:18
It is not quite clear what field this is over, and whether there are maybe additional constraints. Taking the real case at face value, suppose that $,x=y=z=a ne 0,$ for example, then the equation reduces to the quadratic $,(a + 3) w^2 + (3 a^2 + 14 a + 3) w + 3 a^2 + a=0,$, which can certainly have non-zero $,w,$ roots of magnitude other than $1,$.
– dxiv
Jul 15 at 1:18
1
1
@dxiv I'll probably ask a separate question about a system of equations based on this, similar to the one I linked to in the question.
– Shaun
Jul 15 at 1:40
@dxiv I'll probably ask a separate question about a system of equations based on this, similar to the one I linked to in the question.
– Shaun
Jul 15 at 1:40
1
1
Horner Form: $z (x (w (1 + w) + w x) + w x z) + y (x (w (1 + w) + w x) + y (w x + (w + (1 + w) x) z) + z (w (1 + w) + x (1 + w (8 + w) + (1 + w) x) + (w + (1 + w) x) z))$
– Mason
Jul 15 at 1:41
Horner Form: $z (x (w (1 + w) + w x) + w x z) + y (x (w (1 + w) + w x) + y (w x + (w + (1 + w) x) z) + z (w (1 + w) + x (1 + w (8 + w) + (1 + w) x) + (w + (1 + w) x) z))$
– Mason
Jul 15 at 1:41
1
1
There are numerous professional tools for factoring polynomials. Have you tried turning any of them loose on this?
– Steven Stadnicki
Jul 15 at 1:43
There are numerous professional tools for factoring polynomials. Have you tried turning any of them loose on this?
– Steven Stadnicki
Jul 15 at 1:43
6
6
If it were 5wxyz instead of 8wxyz, there would be the factor 1+w+x+y+z
– Empy2
Jul 15 at 3:42
If it were 5wxyz instead of 8wxyz, there would be the factor 1+w+x+y+z
– Empy2
Jul 15 at 3:42
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$$beginalign, p_1(x, y, z, w)=\
& 8wxyz \
&+w^2yz+w^2xz+w^2xy \
&+x^2yz+x^2wz+x^2wy \
&+y^2xz+y^2wz+y^2wx \
&+z^2xy+z^2wy+z^2wx \
&+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \
&+xyz+wyz+wxz+wxytagI
endalign$$
is a polynomial from $mathbbC[x, y, z, w]=mathbbC[x][y][z][w]$. So it can be seen as polynomial $p_2$ of the variable $w$ with coefficients from the integral domain $mathbbC[x][y][z]$.
So we have
$$beginalign, p_2(w)=\
&(yz+xz+xy +xyz)w^2\
&+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)w\
&+(x^2yz
+y^2xz
+z^2xy
+xyz)
endalign$$
The coefficients are from $mathbb C[x,y,z]$. The constant term can be split into the irreducible factors $$x y z (x+y+z+1).$$
The coefficient of $w^2$ is a polynomial of $mathbb C[x,y][z]$:
$$(xy+x+y)z+xy$$
This is linear in $z$. So a factor can only be a common divisor of $xy+x+y$ and $xy$.
But
$$gcd(xy+x+y,xy)=gcd(x+y,xy)=gcd(x^2,x+y)\ gcd(x,x+y)=gcd(x,y)=1$$
So $$yz+xz+xy +xyz$$ is irreducible.
If $p_3(w)$ is a polynomial of degree $gt 0$ that divides $p_2$, then $$p_2(w)=c_1w+c_0,$$ with $c_1, c_0 in mathbb C[x,y,z]$ and
$$c_1 mid (yz+xz+xy +xyz)$$
and $$c_0 mid x^2yz
+y^2xz
+z^2xy
+xyz.$$
So the possible values for $c_1$ are $$a, a(yz+xz+xy +xyz)$$ where $a in mathbb C$ and the possible values for $c_0$ are $$b, bx, bxy, bxyz,b(1+x+y+z), b(1+x+y+z)x, b(1+x+y+z)xy,b(1+x+y+z)xyz$$ where $b in mathbb C.$ Other divisors like $b y$ we will ignore here for symmetry reasons.
So if we check $$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$ we multiply $(2)$ by $$(b(1+x+y+z))^2$$
and get
$$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$
$$beginalign, \
(yz+xz+xy +xyz)(b(1+x+y+z))^2\
+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)\a(yz+xz+xy +xyz)b(1+x+y+z)\
+(x^2yz
+y^2xz
+z^2xy
+xyz)(a(yz+xz+xy +xyz))^2=0 tag3
endalign$$
and we have to check if there are appropriate $a$ and $b$ such that this equality holds.
This task is simpler if we substitute $x,y,z$ by numbers and solve the resulting equations. I did this using a CAS and found out that all possible combinations result in $$a=b=0$$
So there are no proper factors.
answered Aug 14 at 18:05
miracle173
7,09422246
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$beginalign, p_1(x, y, z, w)=\
& 8wxyz \
&+w^2yz+w^2xz+w^2xy \
&+x^2yz+x^2wz+x^2wy \
&+y^2xz+y^2wz+y^2wx \
&+z^2xy+z^2wy+z^2wx \
&+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \
&+xyz+wyz+wxz+wxytagI
endalign$$
is a polynomial from $mathbbC[x, y, z, w]=mathbbC[x][y][z][w]$. So it can be seen as polynomial $p_2$ of the variable $w$ with coefficients from the integral domain $mathbbC[x][y][z]$.
So we have
$$beginalign, p_2(w)=\
&(yz+xz+xy +xyz)w^2\
&+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)w\
&+(x^2yz
+y^2xz
+z^2xy
+xyz)
endalign$$
The coefficients are from $mathbb C[x,y,z]$. The constant term can be split into the irreducible factors $$x y z (x+y+z+1).$$
The coefficient of $w^2$ is a polynomial of $mathbb C[x,y][z]$:
$$(xy+x+y)z+xy$$
This is linear in $z$. So a factor can only be a common divisor of $xy+x+y$ and $xy$.
But
$$gcd(xy+x+y,xy)=gcd(x+y,xy)=gcd(x^2,x+y)\ gcd(x,x+y)=gcd(x,y)=1$$
So $$yz+xz+xy +xyz$$ is irreducible.
If $p_3(w)$ is a polynomial of degree $gt 0$ that divides $p_2$, then $$p_2(w)=c_1w+c_0,$$ with $c_1, c_0 in mathbb C[x,y,z]$ and
$$c_1 mid (yz+xz+xy +xyz)$$
and $$c_0 mid x^2yz
+y^2xz
+z^2xy
+xyz.$$
So the possible values for $c_1$ are $$a, a(yz+xz+xy +xyz)$$ where $a in mathbb C$ and the possible values for $c_0$ are $$b, bx, bxy, bxyz,b(1+x+y+z), b(1+x+y+z)x, b(1+x+y+z)xy,b(1+x+y+z)xyz$$ where $b in mathbb C.$ Other divisors like $b y$ we will ignore here for symmetry reasons.
So if we check $$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$ we multiply $(2)$ by $$(b(1+x+y+z))^2$$
and get
$$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$
$$beginalign, \
(yz+xz+xy +xyz)(b(1+x+y+z))^2\
+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)\a(yz+xz+xy +xyz)b(1+x+y+z)\
+(x^2yz
+y^2xz
+z^2xy
+xyz)(a(yz+xz+xy +xyz))^2=0 tag3
endalign$$
and we have to check if there are appropriate $a$ and $b$ such that this equality holds.
This task is simpler if we substitute $x,y,z$ by numbers and solve the resulting equations. I did this using a CAS and found out that all possible combinations result in $$a=b=0$$
So there are no proper factors.
answered Aug 14 at 18:05
miracle173
7,09422246
add a comment |Â
up vote
1
down vote
accepted
$$beginalign, p_1(x, y, z, w)=\
& 8wxyz \
&+w^2yz+w^2xz+w^2xy \
&+x^2yz+x^2wz+x^2wy \
&+y^2xz+y^2wz+y^2wx \
&+z^2xy+z^2wy+z^2wx \
&+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \
&+xyz+wyz+wxz+wxytagI
endalign$$
is a polynomial from $mathbbC[x, y, z, w]=mathbbC[x][y][z][w]$. So it can be seen as polynomial $p_2$ of the variable $w$ with coefficients from the integral domain $mathbbC[x][y][z]$.
So we have
$$beginalign, p_2(w)=\
&(yz+xz+xy +xyz)w^2\
&+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)w\
&+(x^2yz
+y^2xz
+z^2xy
+xyz)
endalign$$
The coefficients are from $mathbb C[x,y,z]$. The constant term can be split into the irreducible factors $$x y z (x+y+z+1).$$
The coefficient of $w^2$ is a polynomial of $mathbb C[x,y][z]$:
$$(xy+x+y)z+xy$$
This is linear in $z$. So a factor can only be a common divisor of $xy+x+y$ and $xy$.
But
$$gcd(xy+x+y,xy)=gcd(x+y,xy)=gcd(x^2,x+y)\ gcd(x,x+y)=gcd(x,y)=1$$
So $$yz+xz+xy +xyz$$ is irreducible.
If $p_3(w)$ is a polynomial of degree $gt 0$ that divides $p_2$, then $$p_2(w)=c_1w+c_0,$$ with $c_1, c_0 in mathbb C[x,y,z]$ and
$$c_1 mid (yz+xz+xy +xyz)$$
and $$c_0 mid x^2yz
+y^2xz
+z^2xy
+xyz.$$
So the possible values for $c_1$ are $$a, a(yz+xz+xy +xyz)$$ where $a in mathbb C$ and the possible values for $c_0$ are $$b, bx, bxy, bxyz,b(1+x+y+z), b(1+x+y+z)x, b(1+x+y+z)xy,b(1+x+y+z)xyz$$ where $b in mathbb C.$ Other divisors like $b y$ we will ignore here for symmetry reasons.
So if we check $$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$ we multiply $(2)$ by $$(b(1+x+y+z))^2$$
and get
$$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$
$$beginalign, \
(yz+xz+xy +xyz)(b(1+x+y+z))^2\
+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)\a(yz+xz+xy +xyz)b(1+x+y+z)\
+(x^2yz
+y^2xz
+z^2xy
+xyz)(a(yz+xz+xy +xyz))^2=0 tag3
endalign$$
and we have to check if there are appropriate $a$ and $b$ such that this equality holds.
This task is simpler if we substitute $x,y,z$ by numbers and solve the resulting equations. I did this using a CAS and found out that all possible combinations result in $$a=b=0$$
So there are no proper factors.
answered Aug 14 at 18:05
miracle173
7,09422246
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$beginalign, p_1(x, y, z, w)=\
& 8wxyz \
&+w^2yz+w^2xz+w^2xy \
&+x^2yz+x^2wz+x^2wy \
&+y^2xz+y^2wz+y^2wx \
&+z^2xy+z^2wy+z^2wx \
&+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \
&+xyz+wyz+wxz+wxytagI
endalign$$
is a polynomial from $mathbbC[x, y, z, w]=mathbbC[x][y][z][w]$. So it can be seen as polynomial $p_2$ of the variable $w$ with coefficients from the integral domain $mathbbC[x][y][z]$.
So we have
$$beginalign, p_2(w)=\
&(yz+xz+xy +xyz)w^2\
&+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)w\
&+(x^2yz
+y^2xz
+z^2xy
+xyz)
endalign$$
The coefficients are from $mathbb C[x,y,z]$. The constant term can be split into the irreducible factors $$x y z (x+y+z+1).$$
The coefficient of $w^2$ is a polynomial of $mathbb C[x,y][z]$:
$$(xy+x+y)z+xy$$
This is linear in $z$. So a factor can only be a common divisor of $xy+x+y$ and $xy$.
But
$$gcd(xy+x+y,xy)=gcd(x+y,xy)=gcd(x^2,x+y)\ gcd(x,x+y)=gcd(x,y)=1$$
So $$yz+xz+xy +xyz$$ is irreducible.
If $p_3(w)$ is a polynomial of degree $gt 0$ that divides $p_2$, then $$p_2(w)=c_1w+c_0,$$ with $c_1, c_0 in mathbb C[x,y,z]$ and
$$c_1 mid (yz+xz+xy +xyz)$$
and $$c_0 mid x^2yz
+y^2xz
+z^2xy
+xyz.$$
So the possible values for $c_1$ are $$a, a(yz+xz+xy +xyz)$$ where $a in mathbb C$ and the possible values for $c_0$ are $$b, bx, bxy, bxyz,b(1+x+y+z), b(1+x+y+z)x, b(1+x+y+z)xy,b(1+x+y+z)xyz$$ where $b in mathbb C.$ Other divisors like $b y$ we will ignore here for symmetry reasons.
So if we check $$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$ we multiply $(2)$ by $$(b(1+x+y+z))^2$$
and get
$$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$
$$beginalign, \
(yz+xz+xy +xyz)(b(1+x+y+z))^2\
+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)\a(yz+xz+xy +xyz)b(1+x+y+z)\
+(x^2yz
+y^2xz
+z^2xy
+xyz)(a(yz+xz+xy +xyz))^2=0 tag3
endalign$$
and we have to check if there are appropriate $a$ and $b$ such that this equality holds.
This task is simpler if we substitute $x,y,z$ by numbers and solve the resulting equations. I did this using a CAS and found out that all possible combinations result in $$a=b=0$$
So there are no proper factors.
answered Aug 14 at 18:05
miracle173
7,09422246
$$beginalign, p_1(x, y, z, w)=\
& 8wxyz \
&+w^2yz+w^2xz+w^2xy \
&+x^2yz+x^2wz+x^2wy \
&+y^2xz+y^2wz+y^2wx \
&+z^2xy+z^2wy+z^2wx \
&+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \
&+xyz+wyz+wxz+wxytagI
endalign$$
is a polynomial from $mathbbC[x, y, z, w]=mathbbC[x][y][z][w]$. So it can be seen as polynomial $p_2$ of the variable $w$ with coefficients from the integral domain $mathbbC[x][y][z]$.
So we have
$$beginalign, p_2(w)=\
&(yz+xz+xy +xyz)w^2\
&+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)w\
&+(x^2yz
+y^2xz
+z^2xy
+xyz)
endalign$$
The coefficients are from $mathbb C[x,y,z]$. The constant term can be split into the irreducible factors $$x y z (x+y+z+1).$$
The coefficient of $w^2$ is a polynomial of $mathbb C[x,y][z]$:
$$(xy+x+y)z+xy$$
This is linear in $z$. So a factor can only be a common divisor of $xy+x+y$ and $xy$.
But
$$gcd(xy+x+y,xy)=gcd(x+y,xy)=gcd(x^2,x+y)\ gcd(x,x+y)=gcd(x,y)=1$$
So $$yz+xz+xy +xyz$$ is irreducible.
If $p_3(w)$ is a polynomial of degree $gt 0$ that divides $p_2$, then $$p_2(w)=c_1w+c_0,$$ with $c_1, c_0 in mathbb C[x,y,z]$ and
$$c_1 mid (yz+xz+xy +xyz)$$
and $$c_0 mid x^2yz
+y^2xz
+z^2xy
+xyz.$$
So the possible values for $c_1$ are $$a, a(yz+xz+xy +xyz)$$ where $a in mathbb C$ and the possible values for $c_0$ are $$b, bx, bxy, bxyz,b(1+x+y+z), b(1+x+y+z)x, b(1+x+y+z)xy,b(1+x+y+z)xyz$$ where $b in mathbb C.$ Other divisors like $b y$ we will ignore here for symmetry reasons.
So if we check $$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$ we multiply $(2)$ by $$(b(1+x+y+z))^2$$
and get
$$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$
$$beginalign, \
(yz+xz+xy +xyz)(b(1+x+y+z))^2\
+(x^2z+x^2y
+y^2z+y^2x
+z^2y+z^2x
+x^2yz+xy^2z+xyz^2
+yz+xz+xy+8xyz)\a(yz+xz+xy +xyz)b(1+x+y+z)\
+(x^2yz
+y^2xz
+z^2xy
+xyz)(a(yz+xz+xy +xyz))^2=0 tag3
endalign$$
and we have to check if there are appropriate $a$ and $b$ such that this equality holds.
This task is simpler if we substitute $x,y,z$ by numbers and solve the resulting equations. I did this using a CAS and found out that all possible combinations result in $$a=b=0$$
So there are no proper factors.
answered Aug 14 at 18:05
miracle173
7,09422246
answered Aug 14 at 18:05
miracle173
7,09422246
answered Aug 14 at 18:05
miracle173
7,09422246
answered Aug 14 at 18:05
miracle173
7,09422246
7,09422246
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852083%2ffactorise-8wxyzw%25c2%25b2yzw%25c2%25b2xzw%25c2%25b2xyx%25c2%25b2yzx%25c2%25b2wzx%25c2%25b2wyy%25c2%25b2xzy%25c2%25b2wzy%25c2%25b2wxz%25c2%25b2xyz%25c2%25b2wyz%25c2%25b2wxw%25c2%25b2x%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
It is not quite clear what field this is over, and whether there are maybe additional constraints. Taking the real case at face value, suppose that $,x=y=z=a ne 0,$ for example, then the equation reduces to the quadratic $,(a + 3) w^2 + (3 a^2 + 14 a + 3) w + 3 a^2 + a=0,$, which can certainly have non-zero $,w,$ roots of magnitude other than $1,$.
– dxiv
Jul 15 at 1:18
1
@dxiv I'll probably ask a separate question about a system of equations based on this, similar to the one I linked to in the question.
– Shaun
Jul 15 at 1:40
1
Horner Form: $z (x (w (1 + w) + w x) + w x z) + y (x (w (1 + w) + w x) + y (w x + (w + (1 + w) x) z) + z (w (1 + w) + x (1 + w (8 + w) + (1 + w) x) + (w + (1 + w) x) z))$
– Mason
Jul 15 at 1:41
1
There are numerous professional tools for factoring polynomials. Have you tried turning any of them loose on this?
– Steven Stadnicki
Jul 15 at 1:43
6
If it were 5wxyz instead of 8wxyz, there would be the factor 1+w+x+y+z
– Empy2
Jul 15 at 3:42