Question

How do you find all the zeros of `f(x) = x^3 – 3x^2 – 16x + 48`?

Answer 1

`f(x)=x^3-3x^2-16x+48 = color(blue)((x+4)(x-3)(x-4))`
i.e. zeros are at `color(red)(x in {-4,+3,+4})`

Explanation:

Note that a polynomial of degree 3 (such as this one) should have 3 zeroes (although there is no way to be sure that they will all be unique or Real).

There are multiple ways to analyze this but I chose the way that required the least thinking on my part.

Hoping for at least one rational (in this case, integer) zero
I applied the Factor Theorem to all integer factors of `48`
and (using a spreadsheet for the donkey work) got:

This worked even better that I had hoped.
As you can see this gives me all 3 zeros!