Given that $\alpha$ and $\beta$ are roots of the following polynom, show that for $ a \neq 0$
$$( f(x)=ax^2 + bx + c ) \rightarrow f(x)=a(x-\alpha)(x-\beta)$$
I have a question regarding a step in this proof. I'll outline it quickly:
We're given a few theorems to work with. Briefly:
- Division of a polynomial with remainder
- $\alpha$ is a root of $f(x)$ iff $(x-\alpha)|f(x)$
- if $p(x)=q(x)$ then all their coefficients are equal
The first thing we observe is that: $$f(x)=ax^2 + bx + c=h(x)(x-\alpha)(x-\beta)$$
Now here is where I diverge from the proof in the book and have my question. I would open up the right side and observe that:
$$f(x)=h(x)(x^2-(\alpha+\beta)x+\alpha\beta)=h(x)x^2-h(x)(\alpha + \beta)x + h(x)\alpha\beta$$
and according to theorem 3 $h(x)=a$.
However in the book there is an intermediate step in which they prove that $h(x)$ is a constant function. I don't understand why that step is necessary.