Ex 2

I want to lớn factorize the polynomial $x^3+y^3+z^3-3xyz$. Using onfire-bg.comematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?


*

Use Newton"s identities:

$p_3=e_1 p_2 - e_2 p_1 + 3e_3$ and so $p_3-3e_3 =e_1 p_2 - e_2 p_1 = p_1(p_2-e_2)$ as required.

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Here

$p_1= x+y+z = e_1$

$p_2= x^2+y^2+z^2$

$p_3= x^3+y^3+z^3$

$e_2 = xy + xz + yz$

$e_3 = xyz$


*

*

Consider the polynomial $$(lambdomain authority - x)(lambdomain authority - y)(lambdomain authority - z) = lambda^3 - alambda^2+blambda-c ag*1$$We know$$egincasesa = x + y +z\ b = xy + yz + xz \ c = x y zendcases$$ Substitute $x, y, z$ for $lambda$ in $(*1)$ & sum, we get$$x^3 + y^3 + z^3 - a(x^2+y^2+z^2) + b(x+y+z) - 3c = 0$$This is equivalent to$$eginalign x^3+y^3+z^3 - 3xyz= & x^3+y^3+z^3 - 3c\= và a(x^2+y^2+z^2) - b(x+y+z)\= và (x+y+z)(x^2+y^2+z^2 -xy - yz -zx)endalign$$


*

eginalignx^3+y^3+z^3-3xyz & = x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2 \và =(x+y)^3+z^3-3xy(x+y+z)\&= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z) \ và =(x+y+z)(x^2+2xy+y^2+z^2-xy-xz-3xy) \ và =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)endalign


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chú ý that (can be easily seen with rule of Sarrus)$$ eginvmatrix x & y & z \ z và x & y \ y và z & x \ endvmatrix=x^3+y^3+z^3-3xyz$$

On the other hand, it is equal lớn (if we add to the first row 2 other rows)$$ eginvmatrix x+y+z và x+y+z và x+y+z \ z & x & y \ y và z & x \ endvmatrix=(x+y+z)eginvmatrix 1 và 1 & 1 \ z và x và y \ y và z và x \ endvmatrix=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$ just as we wanted. The last echất lượng follows from the expansion of the determinant by first row.


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answered Dec 27 "13 at 15:17
ElensilElensil
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A polynomial from $onfire-bg.combbQ$ is a polynomial from $onfire-bg.combbQ$, so it can be viewed as a polynomial in $z$ with coefficients from the integral domain $onfire-bg.combbQ$.$$p(z)=z^3-3xy cdot z +x^3+y^3$$

So we can try our methods lớn factor a polynomial of degree 3 over an integral domain:If it can be factored then there is a factor of degree $1$, we Hotline it $z-u(x,y)$ và $u(x,y)$ divides the constant term of $p(z)$ which is $x^3+y^3$. The latter is can be factored to lớn $(x+y)(x^2-xy+y^2)$ We check each of the possible values $(x+y), -(x+y), (x^2-xy+y^2), -(x^2-xy+y^2)$ for $u(x,y)$ and find that only $p(-x-y)=0$. So $z-(-x-y)$ is a factor.

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Note:

One can use Kronecker"s method

to lớn reduce the factorization of a polynomial of $onfire-bg.combbQ$ to lớn factoring polynomials in $onfire-bg.combbQ$, lớn reduce the factorization of polynomial of $onfire-bg.combbQ$ to lớn factoring polynomials in $onfire-bg.combbQ$to lớn reduce the factorization of polynomial of $onfire-bg.combbQ$ lớn factoring numbers in $onfire-bg.combbZ$

This factoring is possible in a finite number of steps but the number of steps may become khổng lồ high for practical purpose.

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An integral domain name is a commutative ring with $1$, where the following holds:$$a e 0 lvà b e 0 implies ab e 0$$For polynomials $f$, $g$, $h$ $in I$ this guarantees:$$f=g cdot h implies extdegree(f)= extdegree(g) + extdegree(h) ag1$$compare this to lớn $onfire-bg.combbZ_4$ which is no integral domain và $(2z^2+1)^2 equiv 1$ & so $(2z^n+1) mid 1$. So the polynomial $1$ of degree $0$ has infinitely many divisor.If $I$ is an integral tên miền $(1)$ guarantees that $z^3+az^2+bz+c in I$ has a linear factor và therefore zero in $I$ if it is not irreduzible.