Factorise :

     
I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using maybomnuocchuachay.vnematica 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?


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


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


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

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

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answered Dec 27, 2013 at 15:17
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ElensilElensil
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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.

Here

$p_1= x+y+z = e_1$

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

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

$e_2 = xy + xz + yz$

$e_3 = xyz$


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edited Oct 29, 2013 at 11:31
achille hui
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answered Oct 29, 2013 at 10:29
lhflhf
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A polynomial from $maybomnuocchuachay.vnbbQ$ is a polynomial from $maybomnuocchuachay.vnbbQ$, so it can be viewed as a polynomial in $z$ with coefficients from the integral domain $maybomnuocchuachay.vnbbQ$.$$p(z)=z^3-3xy cdot z +x^3+y^3$$

So we can try our methods khổng lồ factor a polynomial of degree 3 over an integral domain:If it can be factored then there is a factor of degree $1$, we call 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 lớn $(x+y)(x^2-xy+y^2)$ We kiểm tra 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 reduce the factorization of a polynomial of $maybomnuocchuachay.vnbbQ$ lớn factoring polynomials in $maybomnuocchuachay.vnbbQ$, to reduce the factorization of polynomial of $maybomnuocchuachay.vnbbQ$ to lớn factoring polynomials in $maybomnuocchuachay.vnbbQ$to reduce the factorization of polynomial of $maybomnuocchuachay.vnbbQ$ khổng lồ factoring numbers in $maybomnuocchuachay.vnbbZ$

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

Xem thêm: Trường Tiểu Học Tiền Phong A

An integral tên miền is a commutative ring with $1$, where the following holds:$$a e 0 land b e 0 implies ab e 0$$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 $maybomnuocchuachay.vnbbZ_4$ which is no integral domain & $(2z^2+1)^2 equiv 1$ & 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 & therefore zero in $I$ if it is not irreduzible.