Factoring limits in polynomials. Leading coefficient not 1.

Question

I am trying to factor limits in this Mooculus course.

This is the question:

And they have factored it this way:

What is the logic behind factoring it this way? It is not the way where you take two integers with a product of a*c and a sum of b. Nor is it factoring out the initial 2. How would I know that I would use (x+7)?

The answer by the way is:

Thanks

Answer 1

I don't think you're misunderstanding, I think the factoring was just done in a slighty unusual way. Think of it like this:

$$ \begin{align} \frac{2x^2+2x-84}{-2x^2+4x+126}&=\frac{2(x^2+x-42)}{-2(x^2-2x-63)}\\[1ex] &=\frac{2(x-6)(x+7)}{-2(x-9)(x+7)}\\[1ex] &=\frac{(2x-12)(x+7)}{(-2x+18)(x+7)} \end{align} $$

As to why it was written the way it was, it probably was because the only important point was that the two expressions had a common factor of $\,(x+7)\,$.

You also asked how you would know $\,(x+7)\,$ would be a factor of both expressions; that's because otherwise we would not have an indeterminate limit as $\,x \to -7.\,$

Answer 2

After cancelling the common factor $$x+7$$ you can compute the limit $$\lim_{x\to -7}\frac{(2x-12)(x+7)}{-2x+18)(x+7)}=\lim_{x\to -7}\frac{2x-12}{-2x+18}=\frac{2\times (-7)+12}{-2\times (-7)+18}$$