Multivariate factor theorem

The Multivariable Factor Theorem states that If $f(x,y)$ is a polynomial and there is a polynomial $h(x)$ such that $f(x,h(x))=0$ for all $x,$ then we can write $f(x,y) = (y-h(x))g(x,y),$ for some polynomial $g(x,y).$

Proof:

Assume that $f(x, h(x)) = 0$ for all $x$ . We'll treat $x$ as a constant, so that $h(x)$ is constant with respect to $y.$

If we divide $f(x,y)$ by $y-h(x)$ using polynomial long division, so that we have $\underbrace{f(x,y)}_{\textrm{dividend}} = (\underbrace{y - h(x)}_{ \textrm{divisor}}) (\underbrace{q(x,y)}_{\textrm{quotient}}) + \underbrace{r(x)}_{ \textrm{remainder}} \text{ for all } x \text{ and } y.$

Since we're treating $x$ as a constant, $y-h(x)$ is a monic, linear polynomial in $y.$ So, either $r(x)$ is the zero polynomial, in which case it has no terms with $y,$ or it has lower degree in $y$ than $y-h(x).$ This means that $r(x)$ will itself be a polynomial in $x.$

Now, if we set $y=h(x)$ in our equation, it becomes $0 = 0(q(x, h(x))) + r(x) \text{ for all } x.$ It follows that $r(x) = 0.$

So $r(x) = 0$ for any $x,$ and so $r(x)$ is the zero polynomial!

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Multivariate factor theorem