Root of the quadratic equation while solving limit

Question

Let α and β be the distinct roots of $ax^2+bx+c=0$
then $(x-α)(x-β)=0$ is same as $a(x-α)(x-β)=0$
When we solve questions of limit like
$$\lim_{x \to α} \frac{(1−\cos⁡(ax^2+bx+c))}{(x−α)^2}$$
why we write $a(x-α)(x-β)=0$ instead of $(x-α)(x-β)=0$.

Answer 1

To answer the question about why we factor $ax^2+bx+c=0$ as $a(x-\alpha)(x-\beta)=0$ instead of $(x-\alpha)(x-\beta)=0$ is that $(x-\alpha)(x-\beta)$ is not the correct factorization.

If you multiply out $(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta$, we observe that the $x^2$ term doesn't have the right coefficient. We should have $ax^2$ and only have $x^2$. Therefore, if we multiply though by $a$, we'll get the correct factorization.

Basically, $(x-\alpha)(x-\beta)\not=ax^2+bx+c$ because the coefficients can't possibly match (unless $a=1$).

However, $a(x-\alpha)(x-\beta)=0$ has the same zeros and same leading coefficient as $ax^2+bx+c$, so it is the same polynomial.

You can't just simplify the fractions ($\alpha$ and $\beta$ - if they even are fractions) because then you'll end up with coefficients on the $x$'s. For example $$ \left(x-\frac{1}{2}\right)\left(x-\frac{2}{3}\right)=\frac{1}{6}(2x-1)(3x-2). $$

Also, be careful not to confuse the equation $$ ax^2+bx+c=0 $$ with the polynomial $$ ax^2+bx+c. $$ In the equation, you are free to multiply or divide as you choose (since you'll also be multiplying or dividing zero, which doesn't change anything) while with the polynomial, if you multiply by $t$ you must also divide by $t$ to balance.

Answer 2

you must write $$\lim_{x\to \alpha}\frac{(1-\cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2}$$ the result is $$\frac{1}{2}a^2(\alpha-\beta)^2$$ wich can obtained e.g. by L'Hospital.