Writing polynomial as a product of elementary symmetric polynomials












0














Write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 $ as a product of elementary symmetric polynomial



I get $E1=x+y+z$, $E2=xy+xz+yz$, $E3=xyz$.
I've tried factoring out E3(xyz) but I can tell that's not right. I know this probably isn't that difficult, think I'm just going about it the wrong way.



Please help!!










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  • That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
    – José Carlos Santos
    Nov 22 at 19:37










  • why do you have only five terms? you are missing $x^2 z$
    – Will Jagy
    Nov 22 at 19:39










  • @WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
    – 11276
    Nov 22 at 19:54












  • After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
    – Dr. Wolfgang Hintze
    Nov 22 at 19:54












  • @Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
    – 11276
    Nov 22 at 20:01


















0














Write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 $ as a product of elementary symmetric polynomial



I get $E1=x+y+z$, $E2=xy+xz+yz$, $E3=xyz$.
I've tried factoring out E3(xyz) but I can tell that's not right. I know this probably isn't that difficult, think I'm just going about it the wrong way.



Please help!!










share|cite|improve this question
























  • That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
    – José Carlos Santos
    Nov 22 at 19:37










  • why do you have only five terms? you are missing $x^2 z$
    – Will Jagy
    Nov 22 at 19:39










  • @WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
    – 11276
    Nov 22 at 19:54












  • After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
    – Dr. Wolfgang Hintze
    Nov 22 at 19:54












  • @Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
    – 11276
    Nov 22 at 20:01
















0












0








0







Write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 $ as a product of elementary symmetric polynomial



I get $E1=x+y+z$, $E2=xy+xz+yz$, $E3=xyz$.
I've tried factoring out E3(xyz) but I can tell that's not right. I know this probably isn't that difficult, think I'm just going about it the wrong way.



Please help!!










share|cite|improve this question















Write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 $ as a product of elementary symmetric polynomial



I get $E1=x+y+z$, $E2=xy+xz+yz$, $E3=xyz$.
I've tried factoring out E3(xyz) but I can tell that's not right. I know this probably isn't that difficult, think I'm just going about it the wrong way.



Please help!!







abstract-algebra symmetric-polynomials






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 19:53

























asked Nov 22 at 19:32









11276

34




34












  • That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
    – José Carlos Santos
    Nov 22 at 19:37










  • why do you have only five terms? you are missing $x^2 z$
    – Will Jagy
    Nov 22 at 19:39










  • @WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
    – 11276
    Nov 22 at 19:54












  • After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
    – Dr. Wolfgang Hintze
    Nov 22 at 19:54












  • @Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
    – 11276
    Nov 22 at 20:01




















  • That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
    – José Carlos Santos
    Nov 22 at 19:37










  • why do you have only five terms? you are missing $x^2 z$
    – Will Jagy
    Nov 22 at 19:39










  • @WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
    – 11276
    Nov 22 at 19:54












  • After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
    – Dr. Wolfgang Hintze
    Nov 22 at 19:54












  • @Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
    – 11276
    Nov 22 at 20:01


















That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
– José Carlos Santos
Nov 22 at 19:37




That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
– José Carlos Santos
Nov 22 at 19:37












why do you have only five terms? you are missing $x^2 z$
– Will Jagy
Nov 22 at 19:39




why do you have only five terms? you are missing $x^2 z$
– Will Jagy
Nov 22 at 19:39












@WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
– 11276
Nov 22 at 19:54






@WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
– 11276
Nov 22 at 19:54














After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
– Dr. Wolfgang Hintze
Nov 22 at 19:54






After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
– Dr. Wolfgang Hintze
Nov 22 at 19:54














@Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
– 11276
Nov 22 at 20:01






@Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
– 11276
Nov 22 at 20:01












1 Answer
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There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$






share|cite|improve this answer





















  • Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
    – 11276
    Nov 23 at 9:42











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1 Answer
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1 Answer
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active

oldest

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oldest

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There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$






share|cite|improve this answer





















  • Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
    – 11276
    Nov 23 at 9:42
















0














There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$






share|cite|improve this answer





















  • Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
    – 11276
    Nov 23 at 9:42














0












0








0






There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$






share|cite|improve this answer












There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 20:08









Hagen von Eitzen

275k21268495




275k21268495












  • Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
    – 11276
    Nov 23 at 9:42


















  • Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
    – 11276
    Nov 23 at 9:42
















Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
– 11276
Nov 23 at 9:42




Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
– 11276
Nov 23 at 9:42


















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