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On a recent precal test, I saw a question involving the following expression:

$$(x+1)^2-y^2$$

Which factored out into:

$$(x+y+1)(x-y+1)$$

This wasn't very hard, considering that it was already written as the difference of two squares. I then thought "this is easy, but how would I have done it if it weren't easy?"


Let's say that I had to factor something unholy, such as

$$2 a^2 + a^3 - 2 a^4 - a^5 - 2 b - a b + 4 a^2 b - a^4 b - 2 b^2 + a b^2 + 3 a^2 b^2 + 3 a^3 b^2 + a^4 b^2 - 2 b^3 - 3 a b^3 + a^2 b^3 + a^3 b^3 - 2 b^4 - a b^4$$

Of course, this was reverse-engineered, so I already know that the answer is:

$$(a + 2) (a + b + 1) (b - a^2) (a - b^2 - 1)$$

This, in turn, means that the zeros of the polynomial are located at:

$$\begin{align} \ a & = -2 \\ a & = -b-1 \\ b & = a^2 \\ a & = b^2+1 \\ \end{align}$$


One approach that I was thinking about was possibly formatting the polynomial into a table, like so:

$$ \begin{array}{c|lcr} & a^0 & a^1 & a^2 & a^3 & a^4 & a^5 \\ \hline b^0 & 0 & 0 & 2 & 1 & -2 & -1 \\ b^1 & -2 & -1 & 4 & 0 & -1 & 0 \\ b^2 & -2 & 1 & 3 & 3 & 1 & 0 \\ b^3 & -2 & -3 & 1 & 1 & 0 & 0 \\ b^4 & -2 & -1 & 0 & 0 & 0 & 0 \\ \end{array} $$

Does this table tell me anything about the underlying factorization?


Given a large polynomial with two variables, is there a reasonably efficient way of factoring it?

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try this: math.stackexchange.com/a/449556 –  Ahmed Oct 11 '13 at 1:08

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