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February 27th, 2017, 09:56 AM  #1 
Newbie Joined: Feb 2017 From: Peja Posts: 5 Thanks: 0 Math Focus: Jack of all trades, master of none, though often better, than master of one.  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 (wolframalpha) 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 readymade formulas (difference of squares, cubes etc) and grouping. The problem is: $\displaystyle x^2(yz)+y^2(zx)+z^2(xy)$ The solution is: $\displaystyle (xy)(xz)(yz)$ 
February 27th, 2017, 10:46 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,038 Thanks: 423 
The obvious way is to expand first, rearrange, add zero, and then factor. It strikes me as a trick question. $x^2(y  z) + y^2(z  x) + z^2(x  y) =$ $x^2y  x^2z + y^2z  xy^2 + xz^2  yz^2 =$ $x^2y  xy^2 + y^2z  x^2z + xz^2  yz^2 =$ $x^2y  xy^2 + y^2z + (\ xyz + xyz )  x^2z + xz^2  yz^2 =$ $x^2y  xy^2  xyz + y^2z  x^2z + xyz + xz^2  yz^2 =$ $y(x^2  xy  xz + yz)  z(x^2  xy  xz + yz) =$ $(y  z)(x^2  xy  xz + yz)) = (x  y)(x  z)(y  z).$ 
February 27th, 2017, 09:57 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 384 Thanks: 208 Math Focus: Dynamical systems, analytic function theory, numerics 
From inspection you can see that it is symmetric. In particular, if $x = y$ then clearly the polynomial is zero. Thus, $(xy)$ is a factor. Similarly, the polynomial is 0 whenever $y = z$ or $x = z$ implying a factorization of the form $$ x^2(yz) + y^2(zx) + z^2(xy) = (xy)(yz)(zx)p(x,y,z)$$ 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 $(xy)(yz)(zx)$ or equivalently, $(xy)(xz)(yz)$. 
February 28th, 2017, 01:39 PM  #4 
Newbie Joined: Feb 2017 From: Peja Posts: 5 Thanks: 0 Math Focus: Jack of all trades, master of none, though often better, than master of one. 
Thank you both for the replies. SDK, could you please tell me more about your way of solving. So I know that $\displaystyle (xy)$ 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 $\displaystyle (xz)$ 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 $\displaystyle (x−y)(y−z)(z−x)$? 
March 1st, 2017, 06:16 AM  #5  
Senior Member Joined: Sep 2016 From: USA Posts: 384 Thanks: 208 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
If you used the factor $xz$ 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.  
March 1st, 2017, 10:58 AM  #6 
Newbie Joined: Feb 2017 From: Peja Posts: 5 Thanks: 0 Math Focus: Jack of all trades, master of none, though often better, than master of one. 
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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