# Show that $( f(x)=ax^2 + bx + c \wedge a \neq 0 ) \rightarrow f(x)=a(x-\alpha)(x-\beta)$

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:

1. Division of a polynomial with remainder
2. $\alpha$ is a root of $f(x)$ iff $(x-\alpha)|f(x)$
3. 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.

-

Coefficients in a polynomial are constants, by definition. So you can't treat $h(x)$ as a valid coefficient for the purpose of theorem 3 unless you know that it is a constant first.

To elaborate a little bit: $x^2+x\cdot x+1=2x^2+1$, but you can't conclude from this that $1=2$, $x=0$, and $1=1$ (thought the last bit at least is true).

-