# Calculating the coefficients of one form of bi-variate polynomial from its another form

Given a polynomial: $$P(x_1, x_2) = (ax_1+b)(cx_2+d)$$

This can be written in another form as: $$P(x_1, x_2) = d_1x_1x_2 + d_2x_1 + d_3x_2 + d_4$$

where, $d_1 = ac$, $d_2 = ad$, $d_3 = bc$, $d_4 = bd$

A feasible solution does not always exists for the coefficients $a,b,c \text{ and } d$, when $d_1,d_2,d_3 \text{ and } d_4$ are given. This can be seen by checking the condition: $d_1d_4=d_2d_3$.

But, what can be the intuitive explanation of the fact that a feasible solution does not always exists for transformation of two equivalent mathematical expressions?

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Solve the system of equations that you have provided, with known $d_i$ and unknown $a,b,c,d$. – Ilya Aug 27 '12 at 14:08

The problem is in viewing these as two equivalent mathematical expressions. They are not; the set of polynomials of the form $(ax_1+b)(cx_2+d)$ is a proper subset of the set of polynomials of the form $d_1x_1x_2+d_2x_1+d_3x_2+d_4$. I suspect that your sense that it should work the other way has something to do with there being the same number of parameters. For instance, you probably wouldn't find it counterintuitive that every polynomial of the form $(ax_1+b)(cx_2+d)(ex_3+f)$ can be written in the form $d_1x_1x_2x_3+d_2x_1x_3+d_3x_2x_3+d_4x_3+d_5x_1x_2+d_6x_1+d_7x_2+d_8$ but not the other way around, because the one form has $6$ parameters and the other has $8$. But having the same number of parameters isn't a guarantee for being the same set. The expressions $\log x$ and $\exp y$ both have one real parameter, and yet you can write the latter in terms of the former, $\exp y=\log(\exp (\exp y))$, but not the other way around, since $\log x=\exp(\log(\log (x))$ doesn't work if $\log x$ is negative. The number of parameters can be a good heuristic for understanding the "size" of a set, but it often happens that a set is a subset of another set that can be described using the same number of parameters.