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?
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).