Thank you very much for answering my questions!

The document you provided is going to be really useful for me since we are going to be learning more about cubic functions.

If I am not wrong, perhaps the part of the question that says: " State and explain an implication of your result using the Remainder-Factor Theorem." requires some sort of conclusion that a cubic function can always be expressed/factorized to the following 3 forms?

[inspired by a part of the document you provided, so thank you!

Mr F says: You're welcome.]

Since there will always be at least 1 root for a cubic equation,

y= (linear factor)(irreducible quadratic factor)

Mr F says: Yes.
y= (linear factor)^2 (irreducible quadratic factor)

Mr F says: No. (linear factor)^2 gives a quadratic ..... (quadratic)(irreducible quadratic) = quartic, not cubic.
y= (linear factor)(linear factor)(linear factor)

Mr F says: Yes. And included in this case is (linear factor)^2 (linear factor).
Does that mean that once we know a root of the equation, we can always find the factor too?

Mr F says: Yes. If $\displaystyle \alpha$ is a root then $\displaystyle (x - \alpha)$ is always a factor.
(It should be so, right?

Mr F says: Yes.)