If $p(a,b)=0$ where $a,b \in \mathbb R$ then $p(x,y)=(x-a)s(x,y) + (y-b)t(x,y)$ for polynomials $s$ and $t$.

Question

I want to show that if $p$ is a polynomial and $p(a,b)=0$ where $a,b \in \mathbb R$ then $p(x,y)=(x-a)s(x,y) + (y-b)t(x,y)$ for polynomials $s$ and $t$.

This is easy for a single variable polynomial as they are in a Euclidean domain so we can just use the division algorithm, but I am not sure how to do this in general. I would appreciate a hint.

Answer 1

Hint:  let $\,q(x,y)=p(x+a,y+b)\,$, then $\,q(0,0)=0\,$ so $\,q\,$ has no constant term, in other words it is a sum of monomials of degree $\,\ge 1\,$. Take all monomials which contain a power of $\,x\,$ and write their sum as $\,x \cdot q_1(x,y)\,$. The remaining monomials must only contain powers of $\,y\,$ so their sum can be written as $\,y \cdot q_2(y)\,$. So in the end:

$$ \begin{align} q(x,y) = x \cdot q_1(x,y) + y \cdot q_2(y) &\iff p(x+a,y+b) = x\cdot q_1(x,y) + y \cdot q_2(y) \\ &\iff p(x,y) = (x-a) \cdot q_1(x-a,y-b) + (y-b) \cdot q_2(y-b) \end{align} $$

Answer 2

Try reducing to one variable by defining $$p_b(x)=p(x,b).$$

Since $p_b(a)=0$ implies $p_b(x)=(x-a)q_b(x)$. In general, you can write $$p_y(x)=(x-a)q_y(x)+r_y.$$

Now show that both $q(x,y):=q_y(x)$ and $r(y):=r_y$ are polynomials (in two and one variables), and since $r(b)=0$,

$$r(y)=(y-b)u(y).$$

So now $$p(x,y)=(x-a)q(x,y)+(y-b)u(y).$$

This will do.

Answer 3

By writing $x$ as $(x-a)+a$ and $y$ as $(y-b)+b$ we can write $p(x,y)=\sum_{i,j=0}^{N} a_{ij} (x-a)^{i}(y-b)^{j}$. The hypothesis tells us that $c_{00}=0$. Now collect all terms with at least the first power of $(x-a)$. The rest will have $y-b$ as a factor. Done.