The subject line pretty much says it all. In my geometry class today, the following equation came up:
$$x^4 -8a^2x^2 -48a^4 -8bx^3 - 32a^2 bx +16b^2x^2 +64a^2b^2 = 0$$
Specifically, it was in the context of drawing an equilateral triangle inside a square, with one vertex in a corner of the square, and the other two vertices along two of the sides; $x$ is the distance from one vertex of the triangle to the nearest corner of the square, and $a$ and $b$ are constants.
In any case, I'm guessing the above equation is difficult to solve, so factoring, if it's possible, should make it easier. It is possible, by the way, and I have the answer — I can add it, if it helps — but I have no idea how I would go about breaking the polynomial apart.
The only things I notice are the $x^4$ and $-48a^4$ terms; from what I've seen in some other posts, I gather these would be called homogeneous? Must one of the factors have a term containing only some power of $x$? Must one have a term containing only some power of $a$, maybe with a constant attached?
So any guidance would be greatly appreciated. And if you have any suggestions where I could learn about this kind of factoring, that would be terrific too. (I guess by this kind of factoring, I mean anything beyond what's in a normal 1000-level algebra course. I learned that stuff back in tenth grade, something like fifteen years ago, and none of the more advanced classes I've taken recently has covered anything like this.)
Let us denote the proposed polynomial by $F(x,a,b)$.
(1) Note that $F(\lambda x,\lambda a,\lambda b)=\lambda^4F(x,a,b)$, So we may reduce the number of unknowns by choosing $\lambda=\frac{1}{2a}$ for example (the 2 is to reduce the coefficients by the way).
(2) So, we consider $$\eqalign{ F\left(y,\frac{1}{2}, c\right)& = y^4 -8cy^3 - 8 cy +(16c^2-2)y^2 +16c^2-3\cr &=y^4-8cy(y^2+1)+(16c^2-3)y^2 +16c^2-3+y^2\cr &=y^2(y^2+1)-8cy(y^2+1)+(16c^2-3)(y^2 +1)\cr &=(y^2+1)(y^2-8cy +16c^2-3) } $$ and you can continue from here to get:
$$F(x,a,b)=(x-2 i a) (x+2 i a) \left(x-2 \sqrt{3} a-4 b \right) \left(x+2 \sqrt{3} a-4 b \right).$$