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February 27th, 2017, 10:56 AM   #1
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Post Factoring multivariate polynomial

So, I've never been good with factoring polynomials, except maybe quadratic trinomials . Now I have a problem which I know the solution of (wolfram-alpha) but I don't know to how to solve. Now along with asking you to solve that problem for me, I want to know what are some ways you can approach factoring, except the ready-made formulas (difference of squares, cubes etc) and grouping.
The problem is:

$\displaystyle x^2(y-z)+y^2(z-x)+z^2(x-y)$

The solution is:

$\displaystyle (x-y)(x-z)(y-z)$
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February 27th, 2017, 11:46 AM   #2
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The obvious way is to expand first, rearrange, add zero, and then factor. It strikes me as a trick question.

$x^2(y - z) + y^2(z - x) + z^2(x - y) =$

$x^2y - x^2z + y^2z - xy^2 + xz^2 - yz^2 =$

$x^2y - xy^2 + y^2z - x^2z + xz^2 - yz^2 =$

$x^2y - xy^2 + y^2z + (-\ xyz + xyz ) - x^2z + xz^2 - yz^2 =$

$x^2y - xy^2 - xyz + y^2z - x^2z + xyz + xz^2 - yz^2 =$

$y(x^2 - xy - xz + yz) - z(x^2 - xy - xz + yz) =$

$(y - z)(x^2 - xy - xz + yz)) = (x - y)(x - z)(y - z).$
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February 27th, 2017, 10:57 PM   #3
SDK
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From inspection you can see that it is symmetric. In particular, if $x = y$ then clearly the polynomial is zero. Thus, $(x-y)$ is a factor. Similarly, the polynomial is 0 whenever $y = z$ or $x = z$ implying a factorization of the form
$$ x^2(y-z) + y^2(z-x) + z^2(x-y) = (x-y)(y-z)(z-x)p(x,y,z)$$

where $p$ is a polynomial factor to be determined. But you can say even more since the total degree of the cofactor of $p$ is equial to the total degree of the polynomial on the left. Therefore, $p$ is actually a constant.

Finally, you can determine the constant by picking off a single monomial term on the left and comparing it to its counterpart on the right. Multiplication by $p$ must force these to be equal.

For example, on the left there is a $x^2y$ term and by looking on the right we see there is a $-px^2y$ which is enough to conclude that $p = -1$ giving a factorization of $-(x-y)(y-z)(z-x)$ or equivalently, $(x-y)(x-z)(y-z)$.
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February 28th, 2017, 02:39 PM   #4
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Thank you both for the replies.

SDK, could you please tell me more about your way of solving.
So I know that $\displaystyle (x-y)$ is a factor because of the factor theorem, with it being a factor since y is a root (in regard to x).
Could we skip the $\displaystyle p=-1$ part if we just used $\displaystyle (x-z)$ as a factor?
I think this would be allowed, since z is a root (in regard to x).
Also, what happened the $\displaystyle (x,y,z)$ part?
Was it removed because of the degree of the polynomial on the left already being equal to the degree of $\displaystyle (x−y)(y−z)(z−x)$?
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March 1st, 2017, 07:16 AM   #5
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Quote:
Originally Posted by granitba View Post
Thank you both for the replies.

SDK, could you please tell me more about your way of solving.
So I know that $\displaystyle (x-y)$ is a factor because of the factor theorem, with it being a factor since y is a root (in regard to x).
Could we skip the $\displaystyle p=-1$ part if we just used $\displaystyle (x-z)$ as a factor?
I think this would be allowed, since z is a root (in regard to x).
Also, what happened the $\displaystyle (x,y,z)$ part?
Was it removed because of the degree of the polynomial on the left already being equal to the degree of $\displaystyle (x−y)(y−z)(z−x)$?
Yes the factor theorem is what tells you that $(x-y)$ is a factor as well as $(y-z)$ and $(z-x)$. These all follows because of the symmetry in the variables.


If you used the factor $x-z$ you would still have to continue the analysis. However, in this case 2 monomial terms will be the same so when multiplying by $p$ it should not change anything which implies $p = 1$.

Also, initially I wrote $p(x,y,z)$ to denote that $p$ was an arbitrary polynomial in all 3 variables. However, you quickly deduce that it is a constant polynomial so the dependence on $x,y,z$ is omitted. Its not wrong to still write $p(x,y,z) = c$ for some constant $c$ but that is tedious. In fact, the dependence is often not written at all when the context makes it clear but I included it since I don't know what level of math you are doing.
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March 1st, 2017, 11:58 AM   #6
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I do get it now, thanks a lot.
But I'll probably still use Jeff's way of solving most polynomials, and rely to your way if I can't figure it out with the first way.
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