How to show that a root of a bivariate homogeneous polynomial divides the polynomial? For univariate polynomials, one has by polynomial long division that, given any root $a$ of a univariate polynomial $Q(x)$, $(x-a)$ divides $Q$, i.e. $Q(x)=(x-a)f(x)$ for some polynomial $f(x)$.
Now, as part of the definition of root for a bivariate homogeneous polynomial in (an early draft version of) Algebraic Geometry: A Problem Solving Approach, it is claimed in the definition, but not shown, that, given any root $(a,b)$ of the bivariate polynomial $P(x,y)$, $(bx-ay)$ divides $P(x,y)$.
Question: How does one show that $$(a,b)\text{ is a root of }P(x,y) \implies P(x,y)=(bx-ay)f(x,y)?$$
My attempt: First, because in the text one is working with homogeneous polynomials meant to act on points of $\mathbb{CP}^1$ and later $\mathbb{CP}^2$, we can assume that at least one of $a$ and $b$ is not $0$, since there is no point corresponding to $(0,0)$ in the complex projective line.
Now basically I tried to reduce the bivariate case to the univariate case via "de-homogenizing" i.e. evaluating at a given value of $x$ or $y$. In other words, given that $(a,b)$ is a root of $P(x,y)$, it follows immediately that $a$ is a root of $P(x,b)$, so by the result for univariate polynomials, we have $$P(x,b)=(x-a)g(x,b) = (bx-ab)\frac{g(x,b)}{b} $$ assuming that $b\not=0$. Analogously, in the case that $a\not=0$, we have that $$P(a,y)= (y-b)h(a,y) = (ba-ay)\frac{-h(a,y)}{a}.$$ These results together certainly suggest that $(bx-ay)$ divides $P(x,y)$, but I am unsure about at least two things (which are probably dumb misunderstandings on my part, I apologize in advance):
1. How do I argue this when one of $a$ and $b$ is zero? For example, if $b=0$, and thus $bx-ay=-ay$, do I just ignore trying to show that $x-a$ divides $P(x,b)$ and focus only on showing that $y$ divides $P(a,y)$? If not, what's the workaround?
2. How do I show that $f(x,b)=\frac{g(x,b)}{b}$ and that $f(a,y)=\frac{-h(x,a)}{a}$? Do I need to appeal to the fact that we are working in projective space, i.e. results that differ by a non-zero multiplicative factor are equivalent in a standard sense? Or is the result that $(bx-ay)$ divides $P(x,y)$ more generally true?
 A: The valuable concept in this problem is that you can reduce this to the case you already understand, univariate polynomials, by picking an appropriate affine coordinate patch on $\mathbb{P}^1$. If $\mathbb{P}^1=\operatorname{Proj} k[x,y]$, take the affine coordinate patch given by $D(x)$ if $a\neq 0$ and $D(y)$ if $b\neq 0$ (if $a,b\neq 0$, it doesn't matter which one you pick).
Without loss of generality, let's say we picked the patch $D(x)$. The affine coordinate ring of this is $k[y/x]$, and the coordinate of the point which was $(a,b)$ is now $\frac{b}{a}$. On this patch, our polynomial $P(x,y)=\sum a_i x^iy^{d-i}$ is represented by $\widetilde{P}(\frac{y}{x})=\sum a_i (\frac{y}{x})^{d-i}$ (observe this is just $\frac{1}{x^d}P(x,y)$ where $d$ is the degree of $P$). This polynomial has $\frac{b}{a}$ as a root by assumption, which means that $\sum a_i (\frac{y}{x})^{d-i}=(\frac{y}{x}-\frac{b}{a})\widetilde{Q}(\frac{y}{x})$. Multiplying through by $x^d$, we see that $$x^d\widetilde{P}(\frac{y}{x})=x(\frac{y}{x}-\frac{b}{a})x^{d-1}\widetilde{Q}(\frac{y}{x})$$
rearranging, we have 
$$P(x,y)=\frac{1}{a}(ay-bx)Q(x,y)$$
and so we have the desired divisibility relation (remember, $a\neq 0$ by assumption).
