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

May 11, 2016

$f \left(x\right) = {x}^{3} - 3 {x}^{2} - 16 x + 48 = \textcolor{b l u e}{\left(x + 4\right) \left(x - 3\right) \left(x - 4\right)}$
i.e. zeros are at $\textcolor{red}{x \in \left\{- 4 , + 3 , + 4\right\}}$

#### 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!